Notation 2.2. Given a morphism $p:Y \rightarrow X$ of adequate triples, we write $p_{\mathrm {in}}\colon Y_{\mathrm {in}} \rightarrow X_{\mathrm {in}}$ and $p^{\mathrm {eg}}\colon Y^{\mathrm {eg}} \rightarrow X^{\mathrm {eg}}$ for the restriction of p to ingressives and egressives.

Proof. Since the limit of a natural wide subcategory inclusion remains a wide subcategory inclusion, $\big (\lim \limits X, \lim \limits X_{\mathrm {in}}, \lim \limits X^{\mathrm {eg}}\big )$ is indeed a triple of a category together with two wide subcategories (and likewise for filtered colimits). Notice that an ambigressive square in $\lim X$ is cartesian if and only if its image in each $X_i$ is cartesian. Furthermore, a cospan $x\rightarrowtail x'\twoheadleftarrow x"$ in $\lim X$ admits an ambigressive pullback if its images in each $X_i$ do (in this case, those pullbacks assemble into an object of the limit). This implies that $\lim X$ is an adequate triple and that a map $Z\longrightarrow \lim X$ preserves ambigressive pullback squares if and only each composite $Z\longrightarrow \lim X\longrightarrow X_i$ does. This implies that the (pointwise) limit in $\mathrm {Fun} (\Lambda _2[2], \mathrm {Cat} )$ also provides the limit in the subcategory $\mathrm {AdTrip}\hookrightarrow \mathrm {Fun} (\Lambda _2[2], \mathrm {Cat} )$ .

A dual argument applies to the colimit of a filtered diagram. In this case, the image of an ambigressive pullback square in some $X_i$ under the map remains an ambigressive pullback (since at the level of mapping $\infty $ -groupoids, filtered colimits commute with pullbacks). Conversely, any cospan $x\rightarrowtail x'\twoheadleftarrow x"$ in arises as the image of a cospan in some $X_i$ , whose pullback in $X_i$ then provides the desired ambigressive pullback in . It follows that is an adequate triple and that a map of triples preserves ambigressive pullback squares if and only if each does. This implies that is also the colimit in the subcategory $\mathrm {AdTrip}\hookrightarrow \mathrm {Fun} (\Lambda _2[2], \mathrm {Cat} )$ .

Proof. The pointwise pullback P of an ingressive and an egressive natural transformation $F \twoheadrightarrow H \leftarrowtail G$ exists by assumption on Y. To verify that P is indeed a map of adequate triples, note that for an ingressive $x \rightarrowtail x'$ , the map $P(x) \rightarrow P(x')$ participates in the commutative cube

whose front and back face are cartesian by definition of P and whose bottom face is cartesian since $F \twoheadrightarrow H$ is egressive. Consequently, also the top face is cartesian showing that $P(x) \rightarrow P(x')$ is also ingressive. The argument for the preservation of egressives is the same, and that P preserves ambigressives pullbacks follows directly from the pasting laws.

It is also obvious that the maps $P \rightarrow G$ and $P \rightarrow F$ are pointwise egressive and ingressive, respectively, and the argument above also verifies the second condition in the definition of an egressive transformation, and the argument is dual for the ingressives. In total, this shows that $\mathrm {Fun} _{\mathrm {AdTrip}}(X,Y)$ is an adequate triple.

To verify the universal property, let Z be a third adequate triple. In suffices to verify that under the equivalence ${\mathrm {Hom}}_{\mathrm {Cat} }(Z\times X, Y)\simeq {\mathrm {Hom}}_{\mathrm {Cat} }(Z, \mathrm {Fun} (X, Y))$ , a functor $F\colon Z\times X\longrightarrow Y$ is a map of adequate triples if and only if the corresponding map $F'\colon Z\longrightarrow \mathrm {Fun} (X, Y)$ takes values in $\mathrm {Fun} _{\mathrm {AdTrip}}(X, Y)$ and determines a map of adequate triples. To see this, note that any ambigressive square in $Z\times X$ is uniquely a composite of four types of squares:

Now, F preserving ambigressive (cartesian) squares of the first type is equivalent to $F'$ taking values in $\mathrm {Fun} _{\mathrm {AdTrip}}(X, Y)$ . Preserving the ambigressive (automatically cartesian) squares as in the lower row then corresponds precisely to $F'\colon Z\longrightarrow \mathrm {Fun} _{\mathrm {AdTrip}}(X, Y)$ preserving ingressive and egressive morphisms. Finally, F preserves the cartesian ambigressive squares as in the upper right corner if and only if $F'$ preserves ambigressive pullbacks.

Proof. Consider the left solid cospan in the diagram

with $x_1\rightarrowtail x_0\twoheadleftarrow x_2$ its image in X under p. By assumption, the right diagram can be completed to an ambigressive pullback in X, and we let $y_3\rightarrowtail y_2$ be a p-cartesian lift of $x_1\times _{x_0} x_2 \rightarrowtail x_2$ . The universal property of p-cartesian maps then implies that there exists a unique map $y_3\rightarrow y_1$ which lives over $x_1\times _{x_0} x_2\to x_1$ and makes the left square commute. Note that the left square is ambigressive by definition. To see that it is also a pullback, compute

$$ \begin{align*} &\ {\mathrm{Hom}}_Y(y,y_1) \times_{{\mathrm{Hom}}_Y(y,y_0)}{\mathrm{Hom}}_Y(y,y_2) \\ \simeq&\ {\mathrm{Hom}}_X(p(y),x_1) \times_{{\mathrm{Hom}}_X(p(y),x_0)} {\mathrm{Hom}}_Y(y,y_0)\times_{{\mathrm{Hom}}_Y(y,y_0)} {\mathrm{Hom}}_Y(y,y_2) \\ \simeq&\ {\mathrm{Hom}}_X(p(y),x_1) \times_{{\mathrm{Hom}}_X(p(y),x_0)} {\mathrm{Hom}}_Y(p(y),x_2)\times_{{\mathrm{Hom}}_X(p(y),x_2)} {\mathrm{Hom}}_Y(y,y_2) \\ \simeq&\ {\mathrm{Hom}}_X(p(y),x_1 \times_{x_0} x_2) \times_{{\mathrm{Hom}}_X(p(y),x_2)} {\mathrm{Hom}}_Y(y,y_2) \\ \simeq&\ {\mathrm{Hom}}_Y(y,y_3) \end{align*} $$

for $y \in Y$ , where the first equivalence uses the universal property of the cartesian edge $y_1 \rightarrowtail y_0$ and the last one that of $y_3 \rightarrowtail y_2$ . Finally, note that the uniqueness of pullbacks implies that every ambigressive pullback is of this form and is thus clearly preserved by p, which is thus a map of adequate triples.

Proof. Consider a cospan where the left arrow is ingressive and the right arrow is egressive in Y. To show that this admits an ambigressive pullback, note that f decomposes into two egressive maps in Y, one contained in a fibre of p and the other p-cartesian (and likewise for $y_2\twoheadrightarrow y_0$ ). By the pasting lemma for pullbacks, it therefore suffices to show the existence of an ambigressive pullback in the arising special cases.

First, suppose that both f and g are contained in a single fibre $Y_x\simeq F(x)$ . Since the fibre forms an adequate triple, there exists an ambigressive pullback of f and g within $Y_x$ . This square remains a pullback square in Y by [Reference LurieLu09a, Proposition 4.3.1.10] and stays ambigressive by definition.

Next, suppose that g is a p-cartesian lift of an egressive arrow. Then Proposition 2.6 implies the existence of a pullback

that maps to an ambigressive pullback in X. It also implies that the left vertical arrow is again p-cartesian and that the upper horizontal one is p-cartesian or fibrewise if f is (the former by switching the roles of ingressives and egressives in 2.6). In both cases, it follows that the square is ambigressive.

The final case (i.e., if f is assumed a p-cartesian lift of an ingressive arrow) follows in the same fashion by again reversing the roles of the egressives and ingressives in Proposition 2.6.

Construction 2.10. Given a functor $C:D\rightarrow A$ between $\infty $ -categories, we obtain a ‘singular complex’ functor $\mathrm {S}_C \colon A \rightarrow {\mathcal {P}}(D)$ with ${\mathcal {P}}(D) = \mathrm {Fun} (D^{\mathrm {op}},\mathrm {Gpd})$ by currying the composition

$$\begin{align*}D^{\mathrm{op}} \times A \xrightarrow{C^{\mathrm{op}}\times {\mathrm{id}}} A^{\mathrm{op}}\times A \xrightarrow{{\mathrm{Hom}}_{A}} \mathrm{Gpd};\end{align*}$$

that is, $\mathrm {S}_C(X)_d = {\mathrm {Hom}}_A(C(d),X)$ . If A is cocomplete, then $\mathrm {S}_C$ is right adjoint to the colimit extension $|{-}|_C:{\mathcal {P}}(D) \rightarrow A$ ; see [Reference LurieLu09a, Proposition 5.2.6.3].

For the cosimplicial object ${\mathbf {\Delta }} \rightarrow \mathrm {Cat} , n \mapsto [n]$ , we obtain in this fashion an adjunction

where $\mathrm {s}\mathrm {Gpd} = {\mathcal {P}}({\mathbf {\Delta }})$ is the $\infty $ -category of simplicial $\infty $ -groupoids. By results of Joyal, Lurie, Rezk and Tierney [Reference Joyal and TierneyJT07, Reference LurieLu09b, Reference RezkRe01], the nerve functor $\mathrm {N}$ is fully faithful with essential image the full subcategory of complete Segal objects inside $\mathrm {s}\mathrm {Gpd}$ – that is, those simplicial $\infty $ -groupoids T that satisfy the Segal condition and for which the diagram

is cartesian; see [Reference RezkRe10, Section 10] for this characterisation of completeness. A general cosimplicial object C in A therefore gives rise to a functor $A \rightarrow \mathrm {Cat} $ via the composition of $\mathrm {S}_C$ with $\mathrm {ac}$ . When $\mathrm {S}_C$ takes values in (complete) Segal $\infty $ -groupoids, we have a good understanding of this functor, since there is then no need to localise; this happens precisely when the cosimplicial object C satisfies the dual version of the Segal and completeness conditions.

Proof. For each $[n]$ , consider $\operatorname {\mathrm {Tw}}^{r}([n])$ and $\operatorname {\mathrm {Tw}}^{r}([n]^{{\mathrm {op}}})$ with the structure of an adequate triple as in Example 2.9. There is a natural equivalence of cosimplical adequate triples $\operatorname {\mathrm {Tw}}^{r}([-]^{{\mathrm {op}}})\simeq \operatorname {\mathrm {Tw}}^{r}([-])^{\mathrm {rev}}$ , sending an object $(i\geq j)$ in $\operatorname {\mathrm {Tw}}^{r}([n]^{{\mathrm {op}}})$ to the object $(j\leq i)$ in $\operatorname {\mathrm {Tw}}^{r}([n])$ . This induces a natural equivalence of simplicial objects

$$\begin{align*}{\mathrm{Hom}}_{\mathrm{AdTrip}}\big(\operatorname{\mathrm{Tw}}^{r}([-]^{{\mathrm{op}}}),X\big)\simeq {\mathrm{Hom}}_{\mathrm{AdTrip}}\big(\operatorname{\mathrm{Tw}}^{r}([-])^{\mathrm{rev}},X\big)\simeq {\mathrm{Hom}}_{\mathrm{AdTrip}}\big(\operatorname{\mathrm{Tw}}^{r}([-]),X^{\mathrm{rev}}\big) \end{align*}$$

so that the simplicial $\infty $ -groupoid defining $\mathrm {Span}(X)$ is the reverse of that defining $\mathrm {Span}(X^{\mathrm {rev}})$ . The result follows since generally $\mathrm {ac}(T)^{\mathrm {op}} \simeq \mathrm {ac}(T^{\mathrm {rev}})$ as this is true on simplices.

Proof. We shall prove the first claim; the second then follows from 2.14. The triple $(B,A,\iota B)$ is evidently adequate (cf. Example 2.3). Note that the $\infty $ -groupoid of functors from an adequate triple $(X,X_{\mathrm {in}},X^{\mathrm {eg}})$ into $(B,A,\iota B)$ is equivalent to the $\infty $ -groupoid of functors $X\rightarrow B$ which invert the edges in $X^{\mathrm {eg}}$ and take those of $X_{\mathrm {in}}$ into A. Applying this to $X=\operatorname {\mathrm {Tw}}^{r}([n])$ shows that the remaining claim is equivalent to the assertion that the source map

$$\begin{align*}s\colon \operatorname{\mathrm{Tw}}^{r}([n]) \longrightarrow [n]\end{align*}$$

is a localisation (necessarily at those maps whose source component is an equivalence). This is, in fact, true for all $\infty $ -categories C in place of $[n]$ , for example, since $\operatorname {\mathrm {Tw}}^{r}(C) \rightarrow C$ is a cocartesian fibration with contractible fibres (which are always localisations; see Lemma 5.5 below). The localisation property is, however, particularly easy to see when C admits a terminal object. In that case, the source evaluation $\operatorname {\mathrm {Tw}}^{r}(C) \rightarrow C$ is even a Bousfield localisation, with fully faithful left adjoint sending $c \mapsto (c \to \ast )$ .

Proof. Let $\mathcal J_n \subseteq \operatorname {\mathrm {Tw}}^{r}([n])$ denote the subposet consisting of those $(i \leq j)$ with $j \leq i+1$ (i.e., the zig-zag along the bottom). Note that $\mathcal J_n$ decomposes as an iterated pushout

$$\begin{align*}\mathcal J_n \simeq \operatorname{\mathrm{Tw}}^{r}([1]) \cup_{\operatorname{\mathrm{Tw}}^{r}([0])} \operatorname{\mathrm{Tw}}^{r}([1])\cup \dots \cup_{\operatorname{\mathrm{Tw}}^{r}([0])} \operatorname{\mathrm{Tw}}^{r}([1])\end{align*}$$

along the Segal maps; in fact, the nerve $\mathrm {N}(\mathcal J_n)$ is already the iterated pushout of the nerves $\mathrm {N}(\operatorname {\mathrm {Tw}}^{r}([1]))$ in simplicial $\infty $ -groupoids (see [Reference HaugsengHa18, Proposition 5.13] for a similar argument). The iterated pullback appearing in the Segal condition is therefore equivalent to the full subcategory $J_n(X)$ of $\mathrm {Fun} (\mathcal J_n,X)$ spanned by those functors taking left-pointing edges in $\mathcal J_n$ (i.e., those of the form $(i \leq i+1) \rightarrow (i \leq i)$ ) to ingressives and right-pointing arrows, namely, $(i \leq i+1) \rightarrow (i+1 \leq i+1)$ , to egressives. Furthermore, this translates the Segal map to the map ${\mathrm {Q}}_n(X) \rightarrow J_n(X)$ induced by the restriction

$$\begin{align*}\mathrm{Fun} (\operatorname{\mathrm{Tw}}^{r}([n]),X) \longrightarrow \mathrm{Fun} (\mathcal J_n,X).\end{align*}$$

But from the pointwise formula for Kan extensions, one finds that a diagram $F \colon \operatorname {\mathrm {Tw}}^{r}([n]) \rightarrow X$ lies in ${\mathrm {Q}}_n(X)$ if and only if it is right Kan extended from its restriction to $\mathcal J_n$ , which has to lie in $J_n$ . The claim now follows from [Reference LurieLu09a, Proposition 4.3.2.15], since right Kan extension from a full subcategory is fully faithful.

Similarly, for completeness, we first note that the map $P \rightarrow {\mathrm {Q}}_3(X)$ from the pullback in question to the lower left corner is fully faithful since the degeneracy ${\mathrm {Q}}_0(X)^2 \rightarrow {\mathrm {Q}}_1(X)^2$ is (as $|\operatorname {\mathrm {Tw}}^{r}([1])| \simeq *$ ). We claim that its essential image consists exactly of those diagrams whose edges are all equivalences; since also $|\operatorname {\mathrm {Tw}}^{r}([3])| \simeq *$ these are precisely the constant ones, which gives the result. So consider a diagram

all of whose squares are (ambigressive) cartesian and such that the four compositions

$$\begin{align*}F(0 \leq 2) \longrightarrow F(0 \leq 0), \quad F(0 \leq 2) \longrightarrow F(2 \leq 2) \end{align*}$$

$$\begin{align*}F(1 \leq 3) \longrightarrow F(1 \leq 1), \quad F(1 \leq 3) \longrightarrow F(3 \leq 3) \end{align*}$$

are equivalences. Then it first follows that, as pullbacks of equivalences, also $F(0 \leq 3) \rightarrow F(0 \leq 1)$ and $F(0 \leq 3) \rightarrow F(2 \leq 3)$ are equivalences and then by two-out-of-six, the entire outer slopes are. But then by commutativity of the larger rectangles, also $F(0 \leq 1) \rightarrow F(1 \leq 1)$ and $F(2 \leq 3) \rightarrow F(2 \leq 2)$ are equivalences, and then as pullbacks thereof also $F(0 \leq 2) \rightarrow F(1 \leq 2)$ and $F(1 \leq 3) \rightarrow F(1 \leq 2)$ . Finally, this implies that also $F(1 \leq 2) \rightarrow F(1 \leq 1)$ and $F(1 \leq 2) \rightarrow F(2 \leq 2)$ are equivalences by two-out-of-three.

Proof. Let us point out that any colimiting cocone $G\colon I^{\rhd }\to E$ arises as the image of a colimiting cocone $G'\colon I^{\rhd }\to \mathcal {P}(D)$ under $|{-}|_f$ : indeed, one can take $G'$ to be the colimit of the diagram $S_f\circ G_{|I}\colon I\longrightarrow \mathcal {P}(D)$ . Consequently, a functor out of E preserves colimits if and only if its composition with $|{-}|_f$ does. Using this, (1) $\Leftrightarrow $ (2) is an immediate consequence of [Reference LurieLu09a, Lemma 5.1.5.5].

For (2) $\Rightarrow $ (3), (ii) follows immediately. To see (i), note that F agrees with the left Kan extension of $|{-}|_{F\circ f} = F \circ |{-}|_f$ along $|{-}|_f$ , since $|{-}|_f$ is a localisation [Reference LurieLu09a, Proposition 5.2.7.12]. In turn, $|{-}|_{F\circ f}$ is the left Kan extension of $F\circ f$ along the Yoneda embedding $h\colon D\longrightarrow \mathcal {P}(D)$ [Reference LurieLu09a, Lemma 5.1.5.5]. The claim then follows from transitivity of Kan extensions.

Finally, for (3) $\Rightarrow $ (2), note that (ii) implies that $|{-}|_{F\circ f}$ descends along $|{-}|_f$ to some functor $G \colon {E} \rightarrow {C}$ , which is then automatically its left Kan extension along $|{-}|_f$ . Part (i) and transitivity of Kan extensions then imply that $G \simeq F$ , or in other words, that $|{-}|_{F\circ f} \simeq F \circ |{-}|_f$ as desired.

Proof of Proposition 2.20

Since ${\mathrm {Q}}(X)$ is a complete Segal object in $\mathrm {Cat} $ by Lemma 2.17, we have already shown Item (ii) of Lemma 2.21(3). So we need only show that ${\mathrm {Q}}(X) \colon \mathrm {Cat}^{\mathrm {op}} \rightarrow \mathrm {Cat} $ is right Kan extended from its restriction to ${\mathbf {\Delta }}^{\mathrm {op}}$ . By the pointwise formula for Kan extensions, this means that the tautological map

$$\begin{align*}\mathrm{Q}_{D}(X) \longrightarrow \lim_{n \in (\mathbf{\Delta}/D)^{\mathrm{op}}} \mathrm{Q}_{n}(X) \end{align*}$$

is an equivalence for all ${D} \in \mathrm {Cat}$ . The diagram ${\mathbf {\Delta }}/{D} \rightarrow {\mathbf {\Delta }} \hookrightarrow \mathrm {Cat}$ , giving rise to the right-hand side has colimit D and is a typical example of a functor $F \colon I \rightarrow \mathrm {Cat} $ such that the natural map

(*)

is an equivalence (with both sides evaluating to $\mathrm {N} {D}$ ). We will directly show that ${\mathrm {Q}}_{(-)}(X)$ preserves all limits over diagrams with this property.

To this end, note that there are natural equivalences of simplicial $\infty $ -groupoids

$$\begin{align*}\mathrm{N}(\operatorname{\mathrm{Tw}}^{r}({C})) \simeq \big[i \longmapsto {\mathrm{Hom}}_{\mathrm{Cat} }([i] \star [i]^{\mathrm{op}},{C} )\big] \simeq \big[i \longmapsto {\mathrm{Hom}}_{\mathrm{s}\mathrm{Gpd}}(\Delta^i \star (\Delta^i)^{\mathrm{op}},\mathrm{N}({C} ))\big]. \end{align*}$$

In particular, since $\Delta ^i \star (\Delta ^i)^{\mathrm {op}} \simeq \Delta ^{2i+1}$ is completely compact in $\mathrm {s}\mathrm {Gpd}$ , we find an equivalence

because of (*); let us warn the reader that the above map need not be an equivalence without any assumption on F, as the diagram $[1] \xleftarrow {0} [0] \xrightarrow {1} [1]$ with pushout $[2]$ shows. In the case at hand, it follows that the map

is an equivalence.

We have to show that this restricts to an equivalence between the full subcategory on the left (spanned by morphisms of adequate triples) and $\lim _{i \in I}{\mathrm {Q}}_{F_i}(X)$ on the right. To this end, notice that any ambigressive pullback square in $\operatorname {\mathrm {Tw}}^{r}(A)$ , say

is contained (up to equivalence) in $\operatorname {\mathrm {Tw}}^{r}([3])$ for a map $[3]\longrightarrow A$ , namely, that given by the string $a_0 \rightarrow a_1 \rightarrow a_2 \rightarrow a_3$ . Consequently, a functor $\operatorname {\mathrm {Tw}}^{r}(A) \rightarrow X$ is a morphism of adequate triples if and only if each restriction

$$\begin{align*}\operatorname{\mathrm{Tw}}^{r}([3]) \longrightarrow \operatorname{\mathrm{Tw}}^{r}(A) \longrightarrow X\end{align*}$$

is such. To apply this, note that one has

whenever F satisfies (*), so that a functor lies in if and only if for every $j \in I$ and map $[3] \rightarrow F_j$ , the composite

lies in ${\mathrm {Q}}_3(X)$ . But this is the case if and only if the functor restricts to an element of ${\mathrm {Q}}_{F_j}(X)$ for every $j \in I$ and thus by definition if and only if it defines an element of $\lim _{i}{\mathrm {Q}}_{F_i}(X)$ , as desired.

Proof of Theorem 2.18

By Proposition 2.20, $\iota {\mathrm {Q}}(X) \simeq {\mathrm {Hom}}_{\mathrm {AdTrip}}(\operatorname {\mathrm {Tw}}^{r}(-),X) \colon \mathrm {Cat}^{\mathrm {op}} \rightarrow \mathrm {Gpd}$ preserves limits as does ${\mathrm {Hom}}_{\mathrm {Cat} }(-,\mathrm {Span}(X))$ , so by Lemma 2.21, they are both right Kan extended from ${\mathbf {\Delta }}^{\mathrm {op}} \subset \mathrm {Cat}^{\mathrm {op}}$ . Thus, they agree if their restrictions to ${\mathbf {\Delta }}^{\mathrm {op}}$ agree, which is the case by Theorem 2.13.

Proof. The left adjoint $\operatorname {\mathrm {Tw}}^{r}\colon \mathrm {Cat} \rightarrow \mathrm {AdTrip}$ preserves the cartesian product. By adjunction, the right adjoint $\mathrm {Span} \colon \mathrm {AdTrip} \rightarrow \mathrm {Cat} $ then sends the internal mapping object ${\mathrm {Q}}_A(X)=\mathrm {Fun} _{\mathrm {AdTrip}}(\operatorname {\mathrm {Tw}}^{r}(A), X)$ from Lemma 2.5 to the internal mapping object $\mathrm {Fun} (A , \mathrm {Span}(X))$ in $\mathrm {Cat} $ .

Proof. Unwinding definitions, we have to show that for any span in Y, the solid diagram (ignoring the numbers for a moment)

admits an essentially unique dashed filling lying over a given entirely solid diagram in $\mathrm {Span}(X)$ , such that all left-pointing arrows are egressive and all right-pointing arrows ingressive, and the top square is cartesian. We then first observe that the second condition on f (the image of $\psi $ in X) implies that for such square in which the arrows $w \rightarrow u$ and $w \rightarrow \bullet $ are egressive and ingressive, respectively the assertion that the square is cartesian and $\bullet \rightarrow y$ is egressive is equivalent to the assertion that the map $w \rightarrow \bullet $ is p-cocartesian, so we may instead show that there is a unique filler with this property.

We do so by filling the diagram step by step, as indicated by the numbers in the above diagram, essentially uniquely each time:

1. (1) There exists a unique egressive filler because $\phi $ is $p^{\mathrm {eg}}$ -cartesian.

2. (2) The first assumption on f provides a p-cocartesian left, that is automatically unique and also $p_{\mathrm {in}}$ -cocartesian and ingressive.

3. (3) There exists a unique filler making the middle square commute because $w \rightarrow \bullet $ is p-cocartesian.

4. (4) There is a unique ingressive filler because $w\rightarrow \bullet $ is $p_{\mathrm {in}}$ -cocartesian.

Notation 3.6. By forgetting different pieces of the data, the functor $F\colon X^{{\mathrm {op}}}\longrightarrow \mathrm {AdTrip}$ gives three diagrams of $\infty $ -categories $F,F_{\mathrm {in}}$ and $F^{\mathrm {eg}}$ , which are classified by three cartesian fibrations

$$\begin{align*}p\colon Y= \mathrm{Un}^{\mathrm{ct}}(F) \longrightarrow X, \quad p_{\mathrm{in}}\colon Y_{\mathrm{in}} = \mathrm{Un}^{\mathrm{ct}}(F_{\mathrm{in}}) \longrightarrow X \quad \text{and} \quad p^{\mathrm{eg}}\colon Y^{\mathrm{eg}} = \mathrm{Un}^{\mathrm{ct}}(F^{\mathrm{eg}}) \longrightarrow X,\end{align*}$$

respectively, where we generally denote by

$$\begin{align*}\mathrm{Un}^{\mathrm{ct}} \colon \mathrm{Fun} (X^{\mathrm{op}},\mathrm{Cat} ) \longrightarrow \operatorname{\mathrm{Cart}}(X) \quad \text{and} \quad \mathrm{Un}^{\mathrm{cc}} \colon \mathrm{Fun} (X,\mathrm{Cat} ) \longrightarrow \operatorname{\mathrm{Cocart}}(X)\end{align*}$$

the cartesian and cocartesian unstraightening functors, respectively. In our specific situation we further write $Y_{\mathrm {fw}}, Y_{\mathrm {in}}^{\mathrm {fw}},Y^{\mathrm {eg}}_{\mathrm {fw}}$ for the pullback of $Y,Y_{\mathrm {in}},Y^{\mathrm {eg}}$ along $\iota X\rightarrow X$ . Proposition 2.7 then shows that $p\colon Y\longrightarrow X$ is part of a map of adequate triples

which gives rise to a map

(3.7) $$ \begin{align}\mathrm{Span}(p)\colon \mathrm{Span}(Y^{\int})\rightarrow \mathrm{Span}(X,\iota X, X)\simeq X^{{\mathrm{op}}}. \end{align} $$

Proof. Corollary 3.2 immediately implies that $\mathrm {Span}(p)$ is a cocartesian fibration, with cocartesian edges given by spans of the form: where $y\rightarrow y'$ is a $p^{\mathrm {eg}}$ -cartesian edge in Y.

Proof. Consider the full subcategory of $\mathrm {Fun} \big (\mathrm {Fun} (B^{{\mathrm {op}}}, \mathrm {Cat} ), \mathrm {Cat} \big ){/\mathrm {Un}^{\mathrm {ct}}}$ spanned by those natural transformations $\Phi \longrightarrow \mathrm {Un}^{\mathrm {ct}}$ with the following two properties:

1. (1) At the terminal diagram, $\Phi (\ast )\longrightarrow \mathrm {Un}^{\mathrm {ct}}(\ast )\simeq B$ exhibits the inclusion of the core of B.

2. (2) For each $F\colon B^{{\mathrm {op}}}\longrightarrow \mathrm {Cat} $ , the induced square

   (3.14)

   
   is cartesian.

This subcategory is contractible and the natural transformation $\mathrm {Un}^{\mathrm {ct}}(-)\times _B \iota B\longrightarrow \mathrm {Un}^{\mathrm {ct}}(-)$ is (by definition) an object in there. It will therefore suffice to show that the composite natural transformation

satisfies conditions (1) and (2) as well.

Let us start by computing $\Phi (F)$ . To this end, let us write $\operatorname {\mathrm {Ar}}(B)_s=\operatorname {\mathrm {Ar}}(B)\times _{B} \iota B$ , where the structure map $\operatorname {\mathrm {Ar}}(B) \rightarrow B$ is evaluation at the source. Then $t\colon \operatorname {\mathrm {Ar}}(B)_s\longrightarrow B$ is the left fibration classified by the functor $b\mapsto \iota (B{/b})$ and $s\colon \operatorname {\mathrm {Ar}}(B)_s\longrightarrow \iota B$ admits a fully faithful left adjoint $\operatorname {\mathrm {cst}}\colon \iota B\hookrightarrow \operatorname {\mathrm {Ar}}(B)_s$ taking degenerate arrows. Let us now consider the following two pullbacks of $\infty $ -categories

as well as the following diagram of $\infty $ -categories

Since $q\colon C\to \operatorname {\mathrm {Tw}}^{r}(B^{{\mathrm {op}}})$ is the left fibration classifying $\iota (B{/-})\colon \operatorname {\mathrm {Tw}}^{r}(B^{{\mathrm {op}}})\to B\to {\mathrm {Gpd}}$ and $F'$ is constant along the fibres of q, the left Kan extension $q_!F'$ is naturally equivalent to the diagram $F(-)\times \iota (B{/-})$ of which we want to compute the colimit [Reference LurieLu09a, Proposition 4.3.3.10]. By transitivity of Kan extensions, it therefore suffices to compute .

Note that the vertical arrows in the above diagram are all cartesian fibrations, so that the fully faithful inclusion $D\hookrightarrow C$ admits a (localising) right adjoint [Reference LurieLu09a, Corollary 5.2.7.11]

By construction, the diagram $F'$ only depended on the first factor $\operatorname {\mathrm {Tw}}^{r}(B^{{\mathrm {op}}})$ ; Consequently, it factors as $F'= F"p$ , for some $F"\colon D \rightarrow \mathrm {Cat} $ . Because p is a localisation, it is in particular cofinal, so we conclude that .

Let us first compute the left Kan extension $\pi _!F"$ of $F"$ along the cocartesian fibration $\pi \colon D\to \iota B$ . We can compute this Kan extension as a colimit over the fibres, which are equivalently the fibres of $t\colon \operatorname {\mathrm {Tw}}^{r}(B^{{\mathrm {op}}})\to B$ . Unravelling the definitions then shows that

since $B^{{\mathrm {op}}}/b$ has a terminal object. We conclude that there is a natural equivalence

In particular, we see that $\Phi (\ast )$ is an $\infty $ -groupoid, so that (1) $\Phi (\ast )\to B$ is a core inclusion, being a colimit of core inclusions. For (2), note that the vertical maps in (3.14) are cartesian fibrations, so that it suffices to show that the maps on vertical fibres are equivalences. By naturality of the transformation $\Phi \longrightarrow \mathrm {Un}^{\mathrm {ct}}$ in the base B, this map on fibres map is simply the map when $B=*$ . In this case, the map (3.13) is clearly an equivalence.

Proof. The equivalences (3.11) and Lemma 3.12 show that the colimit of $\Psi $ in $\mathrm {Fun} (\Lambda _2[2], \mathrm {Cat} )$ is given by the triple $Y^{\int }$ . To see that this is also a colimit in the subcategory of adequate triples, it suffices to verify that the following are equivalent for a functor :

1. (1) $\phi $ preserves ambigressive pullback squares.

2. (2) the composite does so for each $f^{{\mathrm {op}}}\in \operatorname {\mathrm {Tw}}^{r}(X^{{\mathrm {op}}})$ .

Now note that (on underlying $\infty $ -categories) the map can be identified with the functor $\lambda _f\colon F(x_1)\times X{/x_0}\longrightarrow Y$ sending $(y, g\colon x_2\to x_0)$ to $g^{*}f^{*}(y)$ . Because $g^{*}$ is a map of adequate triples for every map g in X, $\lambda _f$ is a map of adequate triples, which shows that (1) implies (2).

For the converse, note that every ambigressive pullback square in Y can be obtained as a pasting of two types of squares: (a) the pullback of a fibrewise ingressive map along a p-cartesian lift of some $g\colon x_2\to x_0$ and (b) the pullback of an ingressive and egressive map in a single fibre $Y_{x_0}\simeq F(x_0)$ . It suffices to show that each of these squares is the image of an ambigressive pullback square under some $\lambda _f$ . For (a), we can take the image under $\lambda _{\mathrm {id}_{x_0}}$ of the square in $F(x_0)\times X{/x_0}$ formed by an ingressive arrow in $F(x_1)$ and the arrow $g\to \mathrm {id}_{x_0}$ in $X/x_0$ . For (b), we simply take the image under $\lambda _{\mathrm {id}_{x_0}}$ of an ambigressive square in $F(x_0)\times \{\mathrm {id}_{x_0}\}$ .

Proof. Note that F comes with a natural transformation $\iota F\longrightarrow F$ , where each $\iota F(x)$ comes equipped with the (only possible) adequate triple structure where all morphisms are both ingressive and egressive. Unstraightening, this identifies $\mathrm {Un}^{\mathrm {ct}}(\iota F)^{\int }$ with the subcategory of $\mathrm {Un}^{\mathrm {ct}}(F)$ on the cartesian arrows, with all arrows egressive and only the equivalences being ingressive.

The map $\iota F\longrightarrow F$ induces a natural transformation $\iota \mathrm {Span}(F)\simeq \mathrm {Span}(\iota F)\longrightarrow \mathrm {Span}(F)$ , where the first equivalence uses that $\mathrm {Span}(\iota F)\simeq \iota (F)$ (Proposition 2.15). By naturality, we then obtain a commuting diagram

The result now follows since both vertical arrows can be identified with the inclusion of the wide subcategory of cocartesian arrows: for the left arrow this is evident and for the right arrow, this follows from our description of $\mathrm {Un}^{\mathrm {ct}}(\iota F)^{\int }$ and Corollary 3.2.

Proof of Theorem 3.9

We have seen that $\alpha $ constitutes a natural transformation between functors into $\mathrm {Cat} /X^{{\mathrm {op}}}$ . By Lemma 3.16, the value of $\alpha $ at a diagram F is a map between cocartesian fibrations over $X^{\mathrm {op}}$ which preserves cocartesian edges. It then suffices to show that $\alpha _F$ is an equivalence when restricted to every fibre [Reference LurieLu09a, Corollary 2.4.4.4]. Since $\mathrm {Span}$ preserves pullbacks (in particular fibres), we can thus reduce to the case $X=\ast $ , where the map is equivalent to the identity by inspection.

Proof. Note that once we have exhibited the triangle as commutative, the two-out-of-three property will immediately imply that $\mathrm {S}\mathrm {D}^{\mathrm {cc}}$ is an equivalence. We will provide a natural equivalence $\mathrm {S}\mathrm {D}^{\mathrm {cc}}(\mathrm {Un}^{\mathrm {ct}}(F))\longrightarrow \mathrm {Un}^{\mathrm {cc}}(F)$ for functors $F\colon X^{{\mathrm {op}}}\longrightarrow \mathrm {Cat} $ . To this end, consider the functor $\delta \colon \mathrm {Cat} \longrightarrow \mathrm {AdTrip}$ sending $C\mapsto (C, C, \iota C)$ and note that $\mathrm {Span}\circ \delta \simeq \mathrm {id}$ . One then has natural equivalences

$$ \begin{align*} \mathrm{S}\mathrm{D}^{\mathrm{cc}} (\mathrm{Un}^{\mathrm{ct}}(F))\simeq\mathrm{Span}\big(\mathrm{Un}^{\mathrm{ct}}(\delta(F))^{\int}\big)\simeq \mathrm{Un}^{\mathrm{cc}}\big(\mathrm{Span}(\delta(F))\big)\simeq \mathrm{Un}^{\mathrm{cc}}(F), \end{align*} $$

where the middle equivalence is Theorem 3.9.

Construction 3.19. Let us define the lax under-category to be the domain of the cocartesian fibration classifying the functor $(-)^{{\mathrm {op}}}\colon \mathrm {AdTrip}\longrightarrow \mathrm {Cat} $ ; see the next remark for a justification. One can identify an object of with a tuple $(X, x)$ of an adequate triple and an object $x\in X$ , and a map $(X, x)\to (Y, y)$ with a map $f\colon X\to Y$ and a map $\mu \colon y\to f(x)$ in Y. We will say that $(f, \mu )$ is egressive if $\mu $ is egressive, and ingressive if $\mu $ is ingressive and f is an equivalence.

Proof. Apply Theorem 3.9 to the identity functor on $\mathrm {AdTrip}$ and observe that the adequate triple structure from 3.6 corresponds precisely to the one from Construction 3.19 under taking opposite categories.

Proof. The fact that any ambigressive square is cartesian follows immediately from the proof of Proposition 2.6. To show that the ingressives and egressives form a factorisation system on Y, let us start by noting that both classes are stable under retracts (for the cartesian morphisms, this uses that they are characterised by a lifting property relative to the base X). Next, consider a commutative square

where $\phi \in Y^{\mathrm {eg}}$ and $\psi \in Y_{\mathrm {in}}$ . We need to exhibit a unique dotted arrow making the diagram commute. After applying p, there exists a unique dotted edge in X making the square commute. Since $\psi $ is p-cartesian, we find that there is a unique dotted edge in Y which makes the square commute. Finally, given an edge $f\colon x\rightarrow y$ in Y, one can factor $p(f)=gh$ with $g\in X_{\mathrm {in}}$ and $h\in X^{\mathrm {eg}}$ . Taking a p-cartesian lift $g'$ of g, we can factor $f= g'h'$ with $h'\in p^{-1}(X^{\mathrm {eg}})$ , so that $Y^{\dagger }$ and $p^{-1}(X^{\mathrm {eg}})$ form a factorisation system.

Proof. It will suffice to define a natural counit map and provide natural homotopies for the triangle identities. Given an adequate triple X, let $\epsilon _X\colon \operatorname {\mathrm {Ar}}(X)\longrightarrow X$ be the unique functor that fits into a commuting diagram

Here, the left vertical functor is the equivalence from Remark 4.1, a section of which provides a functorial egressive-ingressive factorisation. Since all three solid maps are functorial in the adequate triple X, the map $\epsilon _X$ is functorial in X. Explicitly, the value of $\epsilon _X$ on a map $\mu \colon f\to g$ in $\operatorname {\mathrm {Ar}}(X)$ given by the square

is given by the middle vertical arrow in the unique diagram

From this, we see that $\epsilon _X$ is a map of adequate triples. Indeed, if $\mu $ is ingressive (i.e., $\mu _1$ is an equivalence), then the middle horizontal map is ingressive by the right cancellation property for ingressives [Reference LurieLu09a, Proposition 5.2.8.6 (3)], and likewise in the egressive case where $\mu _0$ is an equivalence.

It remains to provide the triangle identities. If X is an adequate triple, then the composite

is evidently naturally equivalent to the identity (using that the functorial factorisation $\operatorname {\mathrm {Ar}}(X)\to \mathrm {Fact}(X)$ sends degenerate arrows to constant diagrams). Furthermore, for an $\infty $ -category A, the composite

sends an arrow f to $f{\stackrel {\sim }{\longrightarrow }} f$ and then applies the ingressive-egressive factorisation to this natural equivalence. This is again naturally equivalent to the identity.

Proof. To prove the first claim, we use the criterion from Remark 4.1; that is, we show that the map

is an equivalence. We first observe that this map is an equivalence on cores for all adequate triples. This assertion translates to the statement that the $\infty $ -groupoid of dotted extensions of the solid diagram

is contractible, which is obvious.

To conclude, we now observe that it suffices to show that

is an equivalence for all $n \geq 0$ (in fact, $n=1$ suffices, but this will not simplify the argument). But by Corollary 2.22 we can rewrite the target as

$$\begin{align*}{\mathrm{Hom}}_{\mathrm{Cat} }([n] \times [1],\mathrm{Span}(X,X_{\mathrm{in}},X^{\mathrm{eg}})) \simeq \iota\operatorname{\mathrm{Ar}}(\mathrm{Span}({\mathrm{Q}}_n(X,X_{\mathrm{in}},X^{\mathrm{eg}}))).\end{align*}$$

Similarly, the source is equivalent to a set of path components in ${\mathrm {Hom}}_{\mathrm {Cat} }([2],\mathrm {Span}({\mathrm {Q}}_n(X,X_{\mathrm {in}},X^{\mathrm {eg}})))$ , and we claim that it precisely corresponds to $\iota \mathrm {Fact}(\mathrm {Span}({\mathrm {Q}}_n(X,X_{\mathrm {in}},X^{\mathrm {eg}})))$ . Thus, the statement on groupoid cores applied to the various ${\mathrm {Q}}_n(X,X_{\mathrm {in}},X^{\mathrm {eg}})$ gives the statement in full. To verify the remaining claim, one unwinds definitions to find both sides spanned by those diagrams $\operatorname {\mathrm {Tw}}^{r}[n] \times \operatorname {\mathrm {Tw}}^{r}[2] \rightarrow X$ whose restriction along the Segal maps

has all ambigressively marked squares cartesian and all upwards-pointing maps equivalences.

Now assume the adequate triple $(X,X_{\mathrm {in}},X^{\mathrm {eg}})$ is orthogonal. To prove that $(\mathrm {Span}(X), X_{\mathrm {in}}, (X^{\mathrm {eg}})^{{\mathrm {op}}})$ is adequate, we must show that $\mathrm {Span}(X)$ admits ambigressive pullbacks. By Lemma 2.14, we may equivalently show that $\mathrm {Span}(X^{\mathrm {rev}})$ admits ambigressive pushouts, which turns out to be notationally slightly more convenient. Let us start by observing that by adjunction, a commuting square in $\mathrm {Span}(X^{\mathrm {rev}})$ with horizontal arrows in $(X_{\mathrm {in}})^{\mathrm {op}}$ and vertical arrows in $X^{\mathrm {eg}}$ corresponds to a diagram of the form

(*)

in X, including the dotted arrows, whose top right and bottom left squares are pullbacks (so that the arrows with the blue labels are equivalences as well). Such a diagram is uniquely determined by its solid part (i.e., the left column and top row). Indeed, we can first fill the top right square with equivalences in the vertical direction. Then we can uniquely factor the composite in the bottom left square as an egressive followed by an ingressive edge (the resulting square is cartesian because X is an orthogonal adequate triple). Finally, we fill the right two squares with equivalences as indicated.

We can therefore conclude that the triple $(\mathrm {Span}(X), X_{\mathrm {in}}, (X^{\mathrm {eg}})^{\mathrm {op}})$ is both adequate and orthogonal as soon as we can show that any diagram as above defines a pushout square in $\mathrm {Span}(X^{\mathrm {rev}})$ . This is, however, rather unpleasant to do directly. We shall instead use the fact that pushout squares in an $\infty $ -category ${C}$ are precisely the cocartesian edges of the source map $s \colon \operatorname {\mathrm {Ar}}({C}) \rightarrow {C}$ . Here, we regard the diagram ( $\ast $ ) as a morphism in $\operatorname {\mathrm {Ar}}(\mathrm {Span}(X^{\mathrm {rev}}))$ from the left vertical span to the right vertical span, so that the top horizontal span is the image of this morphism under s.

By Corollary 2.22, we can identify the functor $s \colon \operatorname {\mathrm {Ar}}(\mathrm {Span}(X^{\mathrm {rev}})) \rightarrow \mathrm {Span}(X^{\mathrm {rev}})$ with

$$\begin{align*}\mathrm{Span}({\mathrm{Q}}_1X^{\mathrm{rev}}) \xrightarrow{\mathrm{Span}(\mathrm{ev}_{(0 \leq 0)})} \mathrm{Span}(X^{\mathrm{rev}}).\end{align*}$$

Put into this form, we can apply Barwick’s criterion 3.1 for cocartesian edges, or more precisely Corollary 3.2. It tells us that the diagram ( $\ast $ ) defines a $\mathrm {Span}(\mathrm {ev}_{(0 \leq 0)})$ -cocartesian edge if its left-pointing half defines an $(\mathrm {ev}_{(0\leq 0)})^{\mathrm {eg}}$ -cartesian edge in ${\mathrm {Q}}_1X^{\mathrm {rev}}$ .

To see that it does, recall from Lemma 2.5 that an edge in ${\mathrm {Q}}_1X^{\mathrm {rev}} = \mathrm {Fun} _{\mathrm {AdTrip}}(\operatorname {\mathrm {Tw}}^{r}([1]),X^{\mathrm {rev}})$ is egressive if it is pointwise ingressive and its ambigressive square is cartesian. Consider thus the (solid) lifting problem (ignoring the red arrow for a moment)

whose right-hand column is the left half of ( $\ast $ ), and whose lower bent rectangle is a pullback. We have to show that it admits a unique dotted filling (whose lower square is then automatically a pullback by pasting).

Considering the top half of the diagram, there is an essentially unique choice for the upper dotted arrow, because the middle upwards-pointing map is an equivalence, and the composite of ingressives it is itself ingressive. By composition we also obtain the red arrow.

The bottom row of the diagram together with the middle left dot then fit into a diagram

Because the egressives and ingressives in X determine an orthogonal factorisation system, this lifting problem has a unique filler. Moreover, because the right class of an orthogonal factorisation system satisfies right cancellation [Reference LurieLu09a, Proposition 5.2.8.6 (3)], this filler is necessarily ingressive. This provides the necessary lift and concludes the argument.

Observation 4.13. Let us write $\iota {\mathrm {Q}}^{\sharp }\colon \mathrm {s}\mathrm {Gpd}\longrightarrow \mathrm {s}\mathrm {Gpd}$ for the functor sending a simplicial $\infty $ -groupoid S to the simplicial $\infty $ -groupoid given by

$$ \begin{align*} \iota {{\mathrm{Q}}}_n^{\sharp}(S)={\mathrm{Hom}}_{\mathrm{s}\mathrm{Gpd}}\big(\mathrm{N}\operatorname{\mathrm{Tw}}^{r}([n]), S\big). \end{align*} $$

By the definition of span $\infty $ -categories (see Lemma 2.17), there is a natural transformation of functors $\mathrm {AdTrip}\longrightarrow \mathrm {s}\mathrm {Gpd}$

In every simplicial degree, this is given by the inclusion of path components

$$\begin{align*}{\mathrm{Hom}}_{\mathrm{AdTrip}}(\operatorname{\mathrm{Tw}}^{r}([n]),X) \subseteq {\mathrm{Hom}}_{\mathrm{Cat} }(\operatorname{\mathrm{Tw}}^{r}([n]),X).\end{align*}$$

Applying this reasoning twice, one sees that for X in $\mathrm {AdTrip}^{\perp }$ , there is a natural map of simplicial $\infty $ -groupoids

which is an inclusion of path components in each degree. To unravel the target of the above map, let us denote by

$$ \begin{align*} \operatorname{\mathrm{Tw}}^{(2)}(A)=\operatorname{\mathrm{Tw}}^{r}(\operatorname{\mathrm{Tw}}^{r}(A)) \end{align*} $$

the twofold iterated twisted arrow $\infty $ -category of an $\infty $ -category A. We then obtain a natural equivalence

$$ \begin{align*} \iota {\mathrm{Q}}_n^{\sharp}\big(\iota {\mathrm{Q}}^{\sharp}(\mathrm{N} X)\big)&\simeq {\mathrm{Hom}}_{\mathrm{s}\mathrm{Gpd}}\Big(\mathrm{N}\operatorname{\mathrm{Tw}}^{r}([n]), \iota {\mathrm{Q}}^{\sharp}(\mathrm{N} X)\Big)\\ &\simeq \lim_{[m]\in {\mathbf{\Delta}}/\mathrm{N}(\operatorname{\mathrm{Tw}}^{r}[n])} {\mathrm{Hom}}_{\mathrm{s}\mathrm{Gpd}}\big(\mathrm{N}\operatorname{\mathrm{Tw}}^{r}([m]), \mathrm{N} X\big)\\ &\simeq {\mathrm{Hom}}_{\mathrm{s}\mathrm{Gpd}}\big(\mathrm{N}\big(\operatorname{\mathrm{Tw}}^{(2)}([n])\big), \mathrm{N} X\big) \\ &\simeq {\mathrm{Hom}}_{\mathrm{Cat} }(\operatorname{\mathrm{Tw}}^{(2)}([n]), X). \end{align*} $$

Here, the third line uses that $\mathrm {N}\operatorname {\mathrm {Tw}}^{r}(-)$ preserves those colimits of $\infty $ -categories that are preserved by the nerve functor (see the proof of Proposition 2.20). Summarising, we see that there is a map of simplicial $\infty $ -groupoids, depending functorially on $X\in \mathrm {AdTrip}^{\perp }$ , which is a degreewise inclusion of path components

(4.14)

Construction 4.15. Note that $\operatorname {\mathrm {Tw}}^{(2)}([n])$ is equivalent to the poset whose objects are tuples $abcd$ with $0\leq a\leq b\leq c\leq d\leq n$ , corresponding to a map $(a\leq d)\longrightarrow (b\leq c)$ in $\operatorname {\mathrm {Tw}}^{r}([n])$ . The partial order is then given by $abcd\leq a'b'c'd'$ when $a \leq a' \leq b' \leq b \leq c \leq c' \leq d' \leq d$ . We define the following four wide subcategories of $\operatorname {\mathrm {Tw}}^{(2)}([n])$ :

1. (1) $\operatorname {\mathrm {Tw}}^{(2)}([n])_1$ is the subcategory spanned by the edges $abcd \rightarrow a'bcd$ .

2. (2) $\operatorname {\mathrm {Tw}}^{(2)}([n])_2$ is the subcategory spanned by the edges $ab'cd \rightarrow abcd$ .

3. (3) $\operatorname {\mathrm {Tw}}^{(2)}([n])_3$ is the subcategory spanned by the edges $abcd \rightarrow abc'd$ .

4. (4) $\operatorname {\mathrm {Tw}}^{(2)}([n])_4$ is the subcategory spanned by the edges $abcd' \rightarrow abcd$ .

Proof. A functor

$$ \begin{align*} f\colon [n]\longrightarrow {\mathrm{Span}^{\perp}}\circ {\mathrm{Span}^{\perp}}(X)\end{align*} $$

corresponds by adjunction to a map $\operatorname {\mathrm {Tw}}^{r}([n])\longrightarrow {\mathrm {Span}^{\perp }}(X)$ of adequate triples – that is, a map of $\infty $ -categories $f'\colon \operatorname {\mathrm {Tw}}^{r}([n])\longrightarrow {\mathrm {Span}^{\perp }}(X)$ such that

$$ \begin{align*} f'(\operatorname{\mathrm{Tw}}^{r}([n])_{\mathrm{in}})\subseteq {\mathrm{Span}^{\perp}}(X)_{\mathrm{in}}\qquad \text{and}\qquad f'(\operatorname{\mathrm{Tw}}^{r}([n])^{\mathrm{eg}})\subseteq {\mathrm{Span}^{\perp}}(X)^{\mathrm{eg}}. \end{align*} $$

Here, we importantly use that every ambigressive square in the target is automatically a pullback; otherwise, the functor $f'$ would have to furthermore preserve ambigressive pullbacks. By Theorem 2.18, the map of $\infty $ -categories underlying $f'$ corresponds itself to a map $\operatorname {\mathrm {Tw}}^{r}(\operatorname {\mathrm {Tw}}^{r}([n]))\longrightarrow X$ of adequate triples; that is, a map $f"\colon \operatorname {\mathrm {Tw}}^{r}(\operatorname {\mathrm {Tw}}^{r}([n]))\longrightarrow X$ such that

$$ \begin{align*} f"\Big(\operatorname{\mathrm{Tw}}^{r}\big(\operatorname{\mathrm{Tw}}^{r}([n])\big)_{\mathrm{in}}\Big)\subseteq X_{\mathrm{in}}\qquad \text{and} \qquad f"\Big(\operatorname{\mathrm{Tw}}^{r}\big(\operatorname{\mathrm{Tw}}^{r}([n])\big)^{\mathrm{eg}}\Big)\subseteq X^{\mathrm{eg}}. \end{align*} $$

Let us now unravel these conditions using the description of $\operatorname {\mathrm {Tw}}^{(2)}([n])$ from Construction 4.15. Note that, on the one hand, $\operatorname {\mathrm {Tw}}^{r}(\operatorname {\mathrm {Tw}}^{r}([n]))_{\mathrm {in}}$ consists of maps of tuples of the form $abcd\leq a'bcd'$ , while $\operatorname {\mathrm {Tw}}^{r}(\operatorname {\mathrm {Tw}}^{r}([n]))^{\mathrm {eg}}$ consists of maps $abcd\leq ab'c'd$ . On the other hand, $\operatorname {\mathrm {Tw}}^{r}(\operatorname {\mathrm {Tw}}^{r}([n])_{\mathrm {in}})$ consists of maps of the form $abcc\leq a'b'cc$ and $\operatorname {\mathrm {Tw}}^{r}(\operatorname {\mathrm {Tw}}^{r}([n])^{\mathrm {eg}})$ consists of maps $aacd\leq aac'd'$ .

Furthermore, ${\mathrm {Span}^{\perp }}(X)_{\mathrm {in}}$ is the subcategory consisting of spans of the form , where the right map is in $X_{\mathrm {in}}$ . Similarly, ${\mathrm {Span}^{\perp }}(X)^{\mathrm {eg}}$ is the subcategory whose morphisms are spans of the form , where the left map is in $X^{\mathrm {eg}}$ . Combining all this, the above conditions translate as follows:

1. (1) Every $abcd\leq a'bcd'$ is mapped to $X_{\mathrm {in}}$ .

2. (2) Every $abcd\leq ab'c'd$ is mapped to $X^{\mathrm {eg}}$ .

3. (3) Every $abcc\leq a'bcc$ is sent to $X_{\mathrm {in}}$ and every $abcc\leq ab'cc$ is sent to an equivalence.

4. (4) Every $aacd\leq aac'd$ is sent to $X^{\mathrm {eg}}$ and every $aacd\leq aacd'$ is sent to an equivalence.

Note that these conditions are certainly implied by the ones from the proposition. Conversely, given these conditions, the ones from the statement follow: for example, the map $abcd\leq abcd'$ fits into a square

The top horizontal arrow is mapped to an equivalence in X by (4), and the vertical arrows are sent to maps over $X^{\mathrm {eg}}$ by (2). In particular, the bottom arrow maps to $X^{\mathrm {eg}}$ , but also to $X_{\mathrm {in}}$ by (1). An edge which is in both the left and right part of an orthogonal factorisation system is an equivalence, and therefore, $abcd\leq abcd'$ maps to an equivalence.

Proof. Since the domain and the target of $\zeta $ are complete Segal $\infty $ -groupoids, $\zeta $ is an equivalence of simplicial $\infty $ -groupoids as soon as it induces an equivalence in simplicial degrees $0$ and $1$ . It therefore suffices to show that the natural transformation (4.18) is an equivalence of triples of $\infty $ -categories for $n=0$ or $n=1$ .

For $n=0$ , this is evident since the domain and codomain of (4.18) are both just a point. For $n=1$ , at the level of the underlying $\infty $ -categories, we have to verify that the functor $z\colon \operatorname {\mathrm {Tw}}^{(2)}([1])\longrightarrow \operatorname {\mathrm {Ar}}([1])$ exhibits $\operatorname {\mathrm {Ar}}([1])$ as the localisation of $\operatorname {\mathrm {Tw}}^{(2)}([1])$ at the maps in $\operatorname {\mathrm {Tw}}^{(2)}([n])_2$ and $\operatorname {\mathrm {Tw}}^{(2)}([n])_4$ . The map z is given by the map

collapsing the two left-pointing arrows. This is manifestly a localisation. Furthermore, under the map z, the noninvertible arrow in $\operatorname {\mathrm {Tw}}^{r}([1])_1$ (i.e., $0111\to 1111$ ) is sent to the ingressive $01\to 11$ and the noninvertible arrow in $\operatorname {\mathrm {Tw}}^{r}([1])_3$ (i.e., $0001\to 0011$ ) is sent to the egressive $00\to 01$ . We conclude that the map $\mathrm {LTw}^{(2)}([1])\longrightarrow \operatorname {\mathrm {Ar}}([1])$ is an equivalence of (adequate) triples, as desired.

Proof of Theorem 4.12

Lemma 4.19 shows that after post-composing with the forgetful functor $U\colon \mathrm {AdTrip}\rightarrow \mathrm {Cat} $ , there is a natural equivalence

It remains to verify that this equivalence on the underlying $\infty $ -categories also identifies the subcategories of ingressives and egressives. But this is a direct consequence of the naturality of $\zeta $ in $X\in \mathrm {AdTrip}^{\perp }$ , since the inclusions ${\mathrm {Span}^{\perp 2}}(X)_{\mathrm {in}}\hookrightarrow {\mathrm {Span}^{\perp 2}}(X)\hookleftarrow {\mathrm {Span}^{\perp 2}}(X)^{\mathrm {eg}}$ arise as the functors underlying the maps of adequate triples

$$ \begin{align*} {\mathrm{Span}^{\perp 2}}(X_{\mathrm{in}},X_{\mathrm{in}},\iota X)\longrightarrow {\mathrm{Span}^{\perp 2}}(X)\longleftarrow {\mathrm{Span}^{\perp 2}}(X^{\mathrm{eg}},\iota X,X^{\mathrm{eg}}) \end{align*} $$

and the same for $X_{\mathrm {in}}\hookrightarrow X\hookleftarrow X^{\mathrm {eg}}$ .

Observation 5.2. Let us make the following remarks:

1. (1) Let $p\colon Y\longrightarrow X$ be an ingressive cartesian fibration. By definition, Y has all p-cartesian lifts over $X_{\mathrm {in}}$ , and $Y_{\mathrm {in}}$ is the wide subcategory on the p-cartesian lifts of ingressive morphisms. Since $Y_{\mathrm {in}}$ and $Y^{\mathrm {eg}}$ form a factorisation system, this determines $Y^{\mathrm {eg}}$ uniquely and we find that Y arises exactly from the construction in Proposition 4.4; in particular, $Y^{\mathrm {eg}} = p^{-1}(X^{\mathrm {eg}})$ .

2. (2) For a map of orthogonal adequate triples $p \colon Y \rightarrow X$ , an ingressive edge g in Y is p-cartesian if and only if it is $p_{\mathrm {in}}$ -cartesian, since lifting an arbitrary map $f \colon x \rightarrow x'$ in an orthogonal adequate triple along an ingressive $g \colon z \rightarrow x'$ is equivalent to lifting $f' \colon w \rightarrow x'$ along g where $f = f'h$ is the factorisation of f into an egressive followed by an ingressive.

3. (3) Ingressive cartesian fibrations are stable under pullbacks. In particular, if $p\colon Y\longrightarrow X$ is an ingressive cartesian fibration and $x\in X$ , then the fibre over x carries the structure of an orthogonal adequate triple $(Y_x, \iota Y_x, Y_x)$ .

Proof. It is generally true that localisations are cofinal (and coinitial). If p is a localisation, it follows straight from the definition of Kan extension as an adjoint to restriction that any functor $F \colon B \rightarrow C$ is right (and left) Kan extended from its restriction along p and thus has the same colimit as $Fp$ . This proves $(1) \Rightarrow (2)$ .

To see that $(2) \Leftrightarrow (3)$ , we observe that the inclusion

$$\begin{align*}p^{-1}(b) \subseteq b/p, \quad a \longrightarrow (a,{\mathrm{id}}_b),\end{align*}$$

where $b/p$ denotes the pullback $b/B \times _{B} A$ along p, admits a right adjoint, given by taking $(a,f \colon b \rightarrow p(a))$ to a p-cartesian lift of f ending at a. Thus, $|p^{-1}(b)| \simeq |b/p|$ , whence the claim follows from Joyal’s cofinality criterion [Reference LurieLu09a, Theorem 4.1.3.1].

To finally see that $(3) \Rightarrow (1)$ , consider for some $\infty $ -category C the functor

$$\begin{align*}p^{*} \colon \mathrm{Fun} (B,C) \longrightarrow \mathrm{Fun}^{\mathrm{w}}(A,C),\end{align*}$$

where the superscript on the right denotes those functors inverting all maps that p inverts. We have to show that $p^{*}$ is an equivalence. We first claim that it has a right adjoint $p_*$ . By [Reference LurieLu09a, 4.3.3.7], this will follow from

$$\begin{align*}b/p \xrightarrow{\mathrm{fgt}} A \xrightarrow{F} C, \end{align*}$$

admitting a limit for every $b \in B$ and $F \colon A \rightarrow C$ that inverts the fibrewise maps. But as we just discussed above, the inclusion $p^{-1}(b) \subseteq b/p$ admits a right adjoint and is thus coinitial, so we may instead consider

$$\begin{align*}p^{-1}(b) \subseteq A \xrightarrow{F} C, \end{align*}$$

which factors through the localisation $p^{-1}(b) \rightarrow |p^{-1}(b)|$ since F inverts all fibrewise maps by assumption. Thus, the assumption $|p^{-1}(b)| \simeq *$ implies that $F(a)$ is a limit of the above diagram for any $a \in p^{-1}(b)$ . It also follows from the formula for the adjoint as a right Kan extension that the unit and counit of the adjunction $(p^{*},p_*)$ are equivalences, as desired.

Proof. According to Corollary 3.18 above, the embedding $\mathcal {L}^{\mathrm {eg}} \colon \mathrm {AdTrip}^{\perp } \rightarrow \operatorname {\mathrm {Cart}}$ from Proposition 5.4 translates ${\mathrm {Span}^{\perp }}$ to the functor that takes a cartesian fibration to its fibrewise opposite (defined via straightening, postcomposing with $(-)^{\mathrm {op}} \colon \mathrm {Cat} \rightarrow \mathrm {Cat} $ , and then unstraightening). Since $\mathrm {Aut}(\mathrm {Cat} ) \simeq \mathrm {C}_2$ by Toën’s theorem [Reference ToënTo05], this latter operation defines a $\mathrm {C}_2$ -action, and hence, so does the former.

Proof. The forgetful functor factors as $\operatorname {\mathrm {Cart}}\longrightarrow \mathrm {AdTrip}^{\perp }\longrightarrow \mathrm {Cat} $ . The first functor has left adjoint $\mathcal {L}^{\mathrm {eg}}$ by Proposition 5.4, and the second has left adjoint $\operatorname {\mathrm {Ar}}$ by Proposition 4.8. The composite precisely sends A to $s\colon \operatorname {\mathrm {Ar}}(A)\longrightarrow A$ , since this has contractible fibres and the associated orthogonal triple structure on $\operatorname {\mathrm {Ar}}(A)$ is precisely that from Example 4.7.

Proof of Proposition 5.3

Let us start by recalling the model for $X[(X^{\mathrm {eg}})^{-1}]$ given by the relative Rezk nerve [Reference Mazel-GeeMG19]. For each $[n]$ , let us write

for the subcategory of all functors $[n]\longrightarrow X$ and natural transformations which are pointwise egressive. The core of this $\infty $ -category is simply ${\mathrm {Hom}}_{\mathrm {Cat} }([n], X)$ . Taking geometric realisations, we then obtain a map of simplicial $\infty $ -groupoids

such that the induced map on associated $\infty $ -categories is the localisation $X\longrightarrow X[(X^{\mathrm {eg}})^{-1}]$ .

We are now going to give a smaller description of $\big |\mathrm {N}^{\mathrm {rel}}(X)\big |$ , which will show in particular that it already satisfies the Segal condition (so that the associated $\infty $ -category is just its completion). To this end, consider the full subcategories

$$ \begin{align*} \mathrm{N}_{\mathrm{in}}^{\mathrm{rel}}(X)(n)\subseteq \mathrm{N}^{\mathrm{rel}}(X)(n) \end{align*} $$

whose objects are functors $[n]\longrightarrow X$ with values in $X_{\mathrm {in}}$ . For each $\alpha \in \mathrm {N}^{\mathrm {rel}}(X)(n)$ , there exists an initial $\tilde {\alpha }$ in $\mathrm {N}_{\mathrm {in}}^{\mathrm {rel}}(X)(n)$ equipped with a map from $\alpha $ , given by the unique factorisation (starting at the end)

It follows that each inclusion $\mathrm {N}_{\mathrm {in}}^{\mathrm {rel}}(X)(n)\hookrightarrow \mathrm {N}^{\mathrm {rel}}(X)(n)$ admits a left adjoint. We thus obtain a diagram of simplicial categories

(5.9)

where the objects on the left and on the right are simplicial $\infty $ -groupoids and the right vertical map is an equivalence, since it is given in each simplicial degree by the geometric realisation of a right adjoint functor.

The simplicial $\infty $ -category $\mathrm {N}_{\mathrm {in}}^{\mathrm {rel}}(X)$ satisfies the Segal conditions. The equivalence $\mathrm {Fun} ([n], X)\simeq \mathrm {Fun} ([1], X)\times _{X}\dots \times _X \mathrm {Fun} ([1], X)$ restricts to an equivalence of $\infty $ -categories

(5.10)

Here, the pullbacks are taken along the source and target functors. The target functor

(5.11)

is both a left fibration (since each $\rightarrowtail \twoheadrightarrow $ fits into a unique ambigressive square) and a right fibration (since ingressive maps can be pulled back along egressive maps). This implies that the pullbacks along t in (5.10) induce pullbacks after taking realisations (see again [Reference SteimleSt21, Reference SteinebrunnerSt22] or the proof of [CDH+20, Theorem 2.5.1] for the more (co)cartesian version of this assertion), so that the equivalent simplicial $\infty $ -groupoids on the right in (5.9) also satisfy the Segal conditions.

We will use this description of $X[(X^{\mathrm {eg}})^{-1}]$ in terms of $\mathrm {N}^{\mathrm {rel}}_{\mathrm {in}}(X)$ to identify its over-categories. To this end, recall that for any Segal $\infty $ -groupoid S and $s\in S(0)$ , the simplicial $\infty $ -groupoid

(5.12) $$ \begin{align} \big(S/s\big)(n) = S([n+1])\times_{S(\{n+1\})}\{s\} \end{align} $$

is again a Segal $\infty $ -groupoid, such that the associated $\infty $ -category $\mathrm {ac}(S/s)\simeq \mathrm {ac}(S)/s$ is a model for the over-category of s [CDH+20, Lemma 2.4.7]. Let us now take an object $x\in X$ and consider the maps of simplicial $\infty $ -groupoids induced by the top row in (5.9)

(5.13)

Here, the middle term is given by first taking (5.12) at the level of simplicial $\infty $ -categories and then taking realisations. Unraveling the definitions, the simplicial $\infty $ -category $\mathrm {N}^{\mathrm {rel}}_{\mathrm {in}}(X)/x$ is given in degree n by the $\infty $ -category with objects $\alpha (0)\rightarrowtail \dots \rightarrowtail \alpha (n)\rightarrowtail x$ and morphisms

Since the ingressives and egressives form a factorisation system, all vertical maps are equivalences. Consequently, the first map in (5.13) is an equivalence (even before taking geometric realisations). In addition, since the target map (5.11) is a Kan fibration, so is the map $\mathrm {N}_{\mathrm {in}}^{\mathrm {rel}}(X)([n+1])\longrightarrow \mathrm {N}_{\mathrm {in}}^{\mathrm {rel}}(X)(\{n+1\})$ (being a composite of its base changes). It follows that the pullbacks (5.12) are preserved under taking classifying $\infty $ -groupoids, so that the second map in (5.13) is an equivalence. Since $\big |\mathrm {N}^{\mathrm {rel}}_{\mathrm {in}}(X)\big |$ has associated $\infty $ -category $X[(X^{\mathrm {eg}})^{-1}]$ , the associated $\infty $ -category of $\big |\mathrm {N}^{\mathrm {rel}}_{\mathrm {in}}(X)\big |/x$ is $X[(X^{\mathrm {eg}})^{-1}]/p(x)$ .

All in all, we have therefore found that for each object $x\in X$ , the composite functor

is an equivalence. Under this equivalence, the functor p is identified with the functor $X/x\longrightarrow X_{\mathrm {in}}/x$ sending each $y\to x$ to the ingressive part $y'\rightarrowtail x$ of its functorial egressive-ingressive factorisation. This functor admits a fully faithful right adjoint, whose essential image is given by the ingressive maps to x. By [Reference Ayala and FrancisAF20, Lemma 2.16], this implies that p is a cartesian fibration and that an arrow in X is p-cartesian if and only if it is ingressive.

In particular, we can apply Proposition 4.4 to obtain an orthogonal triple structure on X whose ingressives are the p-cartesian arrows and whose egressives are the maps that are inverted by p. Since the first class coincides with $X_{\mathrm {in}}$ , the second class coincides with $X^{\mathrm {eg}}$ .