Proof. If H is an edge, then $\eth(G)$ contains a pair $\{a,a^\dagger\} = \eth(f)$ . Since G is connected, it must be an edge as well.

Now suppose $V_H$ is inhabited. The argument from the proof of [ Reference Hackney, Robertson and Yau33 , lemma 1·26] shows that $V_H \to V_G$ is surjective, hence a bijection. Since f is étale, $D_H \to D_G$ is an isomorphism as well. If there were an element $a\in \eth(G) \setminus \eth(f)$ , then since G is not an edge we would have $a^\dagger \in D_G$ , hence $a^\dagger = f(d)$ for some (unique) $d\in D_H$ . But then $f(d^\dagger) = a \in \eth(G)$ , and only elements of $\eth(H)$ can map to elements of $\eth(G)$ , so $d^\dagger \in \eth(H)$ . This contradicts the assumption on a. It follows that $\eth(H) \hookrightarrow \eth(G)$ is also surjective, hence $A_H \to A_G$ is an isomorphism.

Proof. We compute, using that embeddings are injective on vertices:

\begin{align*}\varsigma[h] + \varsigma[k] &= \sum_{u\in V_H} h(u) + \sum_{w\in V_K} k(w) = \sum_{u^{\prime}\in f(V_H)} \ell(u^{\prime}) + \sum_{w^{\prime}\in g(V_K)} \ell(w^{\prime}) \\[5pt] &= \sum_{v\in f(V_H) \cup g(V_K)} \ell(v) + \sum_{v\in f(V_H) \cap g(V_K)} \ell(v) \leq \varsigma[\ell] + \sum_{v\in f(V_H) \cap g(V_K)} \ell(v).\end{align*}

The last inequality is an equality if and only if $f(V_H) \cup g(V_K) = V_L$ , that is, if and only if $\ell$ is a union of h and k. If h and k are vertex disjoint then $f(V_H) \cap g(V_K)$ is empty, so the final statement holds.

Proof. If L has only a single vertex v, then since L contains an edge, we can form

where g classifies some arc of L. The conclusion follows.

Now suppose that L has two or more vertices. Consider the classical graph $\textrm{core}(L)$ obtained by deleting the $\dagger$ -orbit of $\eth(L)$ ([ Reference Hackney, Robertson and Yau33 , definition 1·9]). As $\textrm{core}(L)$ is connected and has more than one vertex, there exists a vertex v so that deleting v and all edges incident to it from $\textrm{core}(L)$ will still give a connected graph. Setting

\begin{align*}S = nb(v)^\dagger \cap (\eth(L)\cup nb(v)),\end{align*}

we have $S^\dagger \subsetneq nb(v)$ ; also notice that $S \cap S^\dagger = nb(v) \cap nb(v)^\dagger$ . Define a graph K by

\begin{align*}V_K &= V_L \setminus \{ v\} \\[5pt] D_K &= D_L \setminus nb(v) \\[5pt] A_K &= A_L \setminus (S \cup S^\dagger).\end{align*}

We compute the boundary of K as

\begin{align*} \eth(K) = (\eth(L) \setminus S) \cup (nb(v) \setminus S^\dagger),\end{align*}

and use this to show that $\eth(K)^\dagger \subseteq D_K$ . If $a\in \eth(L)\setminus S$ , then $a^\dagger \notin nb(v)$ , hence $a^\dagger \in D_K$ . If $a \in nb(v) \setminus S^\dagger$ , then we can’t have $a^\dagger \in \eth(L)$ for then $a^\dagger \in nb(v)^\dagger \cap \eth(L) \subseteq S$ , and we also can’t have $a^\dagger \in nb(v)$ since then $a \in nb(v)\cap nb(v)^\dagger \subseteq S^\dagger$ ; it follows that $a^\dagger\in D_K$ . We have thus shown that K does not contain any edge components. As $\textrm{core}(K) = \textrm{core}(L) - v$ is connected, we conclude that K is connected. The inclusions of subsets give our desired embedding $g\colon K \rightarrowtail L$ .

Proof. If L is an edge, so is $\hat \varphi[\ell]$ by Definition 4·2(i), hence $\varsigma \hat \varphi[\ell] = 0$ as desired. If L is a star, then the formula is immediate. We have thus established the formula whenever L does not contain an internal edge.

For the general result, we induct on the number of vertices of L, which we assume to have an internal edge (in other words, we are investigating what happens when L is not an edge and not a star). Applying Lemma 4·9, we obtain a diagram

with $\ell$ a union of h and k, and h, k vertex disjoint. Applying $\hat \varphi$ , we obtain embeddings

with $\hat \ell$ a union of $\hat h$ and $\hat k$ (by Definition 4·2(ii)) and $\hat h, \hat k$ vertex disjoint (by Definition 4·2(iii)), hence

\begin{align*} \varsigma \hat \varphi [\ell] = \varsigma [\hat \ell] = \varsigma [\hat k] + \varsigma [\hat h] = \varsigma \hat \varphi [k] + \varsigma \hat \varphi [\ell \iota_v] \end{align*}

by Lemma 4·8. As K has one less vertex than L, by the induction hypothesis we have the first equality below:

\begin{align*} \varsigma \hat \varphi [k] = \sum_{w\in V_K} \varsigma \hat \varphi [k \iota_w] = \sum_{w\in V_K} \varsigma \hat \varphi [\ell g \iota_w] = \sum_{u\in V_L \setminus \{ v\} } \varsigma \hat \varphi [\ell \iota_u].\end{align*}

The result follows by combining the previous two displays.

Proof of Theorem 4·7. Suppose $(\varphi_0, \hat \varphi) \colon G \to G'$ is a new graph map. We verify each of the three conditions from Definition 2·9.

1. (i) Applying Proposition 4·10 to the identity embedding $id_G \colon G \rightarrowtail G$ , we find

   \begin{align*} \sum_{v\in V_G} \varsigma(\varphi_1(v)) = \sum_{v\in V_G} \varsigma \hat \varphi[\iota_v] = \varsigma \hat \varphi[id_G] \leq \varsigma [id_{G^{\prime}}] = \sum_{w\in V_{G^{\prime}}} w. \end{align*}

2. (ii) This is the special case of the interpretation of Definition 4·2(iv) from Remark 4·3 where we take $h = \iota_v$ :

   

3. (iii) Suppose the boundary of G is empty, and consider an embedding $h \colon H \rightarrowtail G'$ which represents $\hat \varphi [id_G]$ . By Definition 4·2(iv) we have that $\eth(H)$ is empty, hence h is an isomorphism by Lemma 2·16. By Proposition 4·10 we have

   \begin{align*} \sum_{v\in V_G} \varsigma \varphi_1(v) = \varsigma \hat \varphi [id_G] = \varsigma [h] = \varsigma [id_{G^{\prime}}] = \sum_{w\in V_{G^{\prime}}} w.\end{align*}
   Since $\eth(G')$ is empty, G ^′ contains at least one vertex, implying that there is at least one $v\in V_G$ with $\varsigma \varphi_1(v) \neq 0$ .

Proof. The second statement is clear since we have an active-inert orthogonal factorisation system on $\textbf{U}$ . The first statement also follows from the factorisation system: suppose $\ell \colon L \rightarrowtail G$ is an embedding.

First factor $\varphi \ell$ as $f\circ p$ , and then factor $\psi f$ as $h\circ q$ , and notice that $h \circ (qp)$ is an active-inert factorisation of $\psi \varphi \ell$ . It follows that $\hat \varphi [\ell] = [f]$ , $\hat \psi [f] = [h]$ , and $\widehat {\psi \circ \varphi} [\ell] = [h]$ .

Proof. Suppose $v\in V_{L^{\prime}}$ . Our goal is to show that $\hat \ell (v)$ is an element of $\hat h(V'_{\!\!H}) \cup \hat k(V_{K^{\prime}})$ . Since $\alpha \colon L\to L'$ is active, by [ Reference Hackney, Robertson and Yau33 , proposition 2·3] there is a unique vertex $w\in V_L$ so that $\alpha_w \colon M_w \rightarrowtail L'$ (representing $\alpha_1(w)$ ) has v in its image. As $V_H \amalg V_K \to V_L$ is surjective, there is a vertex w ^′ of H or K mapping to w; without loss of generality we assume $w^{\prime}\in V_H$ . Considering the square

of (4·2), we get that the triangle

commutes by the descriptions of composing with embeddings from [ Reference Hackney, Robertson and Yau33 , definition 1·34]. Hence v is in the image of $H' \rightarrowtail L'$ , so $\hat \ell (v)$ is an element of $\hat h(V_{H^{\prime}})$ .

Proof. We prove the contrapositive. Suppose $\hat h(V_{H^{\prime}}) \cap \hat k(V_{K^{\prime}})$ is inhabited and let $\hat h(v^{\prime}) = \hat k(w^{\prime})$ be a vertex in this intersection. Since $\beta \colon H \to H'$ is active, by [ Reference Hackney, Robertson and Yau33 , proposition 2·3], there exists a unique vertex $v \in V_H$ so that $\beta_{v} \colon M_{v} \rightarrowtail H'$ (with $[\beta_v] = \beta_1(v)$ ) has v ^′ in its image. Likewise, there exists a unique vertex $w\in V_K$ so that $\gamma_w \colon N_w \rightarrowtail K'$ has w ^′ in its image. We have $\alpha_1(f(v)) = [f' \circ \beta_v]$ and $\alpha_1(g(w)) = [g' \circ \gamma_w]$ by remark 1·45 of [ Reference Hackney, Robertson and Yau33 ].

Now f ^′(v ^′) and g ^′(w ^′) are in the images of both $\alpha_1(f(v))$ and $\alpha_1(g(w))$ ; since $\alpha$ is a graphical map, Definition 2·9 (i) gives $f(v) = g(w)$ . Hence $h(v) = \ell f(v) = \ell g(w) = k(w)$ , so h and k are not vertex disjoint.

Proof. We verify the four conditions from Definition 4·2. Condition (i) follows from lemma 2·2 of [ Reference Hackney, Robertson and Yau33 ]. The preceding two lemmas give (ii) and (iii), respectively. The interpretation of (iv) given in Remark 4·3 is lemma 1·47 of [ Reference Hackney, Robertson and Yau33 ].

Proof. The conclusion will follow once we establish the bijection, as Lemma 4·11 implies that $\mathfrak{N}$ is an identity-on-objects functor from $\textbf{U}$ to the category of new graph maps.

It is nearly immediate that $\mathfrak{O}\mathfrak{N}$ is an identity on $\textbf{U}(G,G')$ , by the same reasoning as in the first bullet point of remark 1·45 of [ Reference Hackney, Robertson and Yau33 ]. Namely, if $\varphi_v \colon H_v \rightarrowtail G'$ represents $\varphi_1(v)$ , then we have a factorisation in $\textbf{U}$

so $\mathfrak{O}\mathfrak{N}(\varphi)_1(v) = \hat \varphi([\iota_v]) = [\varphi_v] = \varphi_1(v)$ .

Now suppose that $\varphi = (\varphi_0, \hat \varphi) \colon G \to G'$ is a new graph map, $\mathfrak{O}(\varphi) = (\varphi_0, \varphi_1)$ is the associated graphical map, and $\mathfrak{N}\mathfrak{O}(\varphi) = (\varphi_0, \tilde \varphi)$ , where $\tilde \varphi$ is a function from $\operatorname{Emb}\!(G)$ to $\operatorname{Emb}\!(G')$ . Let $f \colon H \rightarrowtail G$ be an arbitrary embedding; we wish to show that $\tilde \varphi [f] = \hat \varphi [f]$ . Notice that we already have a special case: if the domain of f is a star, then $[f] = [\iota_v]$ for some $v \in V_G$ , and we know $\tilde \varphi [\iota_v] := \mathfrak{O}(\varphi)_1(v) := \hat \varphi [\iota_v]$ . Let us turn the general case.

Since $\varphi$ and $\mathfrak{N}\mathfrak{O}(\varphi)$ are new graph maps, by Remark 4·3, we have the commutative diagram

and we conclude that $\eth(\hat\varphi [f]) = \eth(\tilde\varphi [f])$ . This is nearly enough to conclude that $\hat\varphi [f]$ and $\tilde\varphi [f]$ are equal — the only thing that could go wrong is if one of them is an edge while the other is not. Applying Proposition 4·10 twice we have

\begin{align*} \varsigma \hat \varphi [f] = \sum_{v\in V_H} \varsigma \hat \varphi [f\iota_v] = \sum_{v\in V_H} \varsigma \tilde \varphi[f\iota_v] = \varsigma \tilde \varphi [f]\end{align*}

where the middle equality comes from the previously established fact that $\hat \varphi [f\iota_v] = \tilde \varphi [f\iota_v]$ . Since an embedding h is an edge if and only if $\varsigma[h] = 0 $ , we conclude that $\hat \varphi [f]$ is an edge if and only if $\tilde \varphi [f]$ is an edge. By Lemma 2·15, we conclude that $\hat \varphi[f] = \tilde \varphi[f]$ . Thus $\varphi = \mathfrak{N}\mathfrak{O}(\varphi)$ .

Definition 5·1 (Directed graphs). Let $\mathscr{G}$ denote the category

A directed graph G is a functor $\mathscr{G} \to \textbf{FinSet}$ which sends the two maps with codomain $\texttt{e}$ to monomorphisms, that is, a directed graph is a diagram of finite sets of the form

\begin{align*} E_{G} \hookleftarrow I_{G} \rightarrow V_{G} \leftarrow O_{G} \hookrightarrow E_{G}. \end{align*}

Each vertex $v\in V_G$ has a set of inputs $\operatorname{in}\!(v) \subseteq I_G$ which is the preimage of v under $I_G \to V_G$ , and likewise has a set of outputs $\operatorname{out}\!(v) \subseteq O_G$ . The set of inputs of G is defined to be $\operatorname{in}\!(G) := E_G \setminus O_G$ and the set of outputs of G is $\operatorname{out}\!(G) := E_G \setminus I_G$ .

Definition 5·2. An étale map $f\colon G \to H$ between directed graphs is a commutative diagram of sets

so that the middle two squares are pullbacks. An embedding is an étale map between connected directed graphs which is injective on vertices.

Definition 5·3 (Orientation presheaf). The orientation presheaf, denoted $\mathfrak{o}$ , is the $\textbf{U}$ -presheaf so that

\begin{align*}\mathfrak{o}_G \subseteq \prod_{a\in A_G} \{ +1, -1 \}\end{align*}

consists of those elements $x = (x_a)$ satisfying $x_a = -x_{a^\dagger}$ for all $a\in A_G$ . If $\varphi \colon H \to G$ is a graphical map and $x\in \mathfrak{o}_G$ , then $\varphi^*(x) \in \mathfrak{o}_H$ is the element with

\begin{align*} \varphi^*(x)_a = x_{\varphi_0(a)}\end{align*}

for all $a\in A_H$ .

Definition 5·7 (Inputs and outputs of embeddings). Suppose G is a directed graph. Then there are functions $\operatorname{in} \colon \operatorname{Emb}\!(G) \to \wp(A_G^+) \cong \wp(E_G)$ and $\operatorname{out} \colon \operatorname{Emb}\!(G) \to \wp(A_G^-) \cong \wp(E_G)$ defined by

\begin{align*} \operatorname{in}\!([f]) &= \eth([f]) \cap A_G^+ \\[5pt] \operatorname{out}\!([f]) &= \eth([f]) \cap A_G^-. \end{align*}

Alternatively, we can consider $f\colon (H,f^*x) \rightarrowtail (G,x)$ as an embedding between directed graphs and we have

\begin{align*}\operatorname{in}\!(f) &= f(\!\operatorname{in}\!(H)) \subseteq A_G^+ \\ \operatorname{out}\!(f) &= f(\!\operatorname{out}\!(H)) \subseteq A_G^-.\end{align*}

Definition 5·8 (Oriented graphical category). The oriented graphical category is the category $\textbf{U}_{/\mathfrak{o}} := \textbf{U} \times_{\widehat{\textbf{U}}} \widehat{\textbf{U}}_{/\mathfrak{o}}$ whose objects are pairs consisting of an undirected graph G and map of presheaves $x \colon G \to \mathfrak{o}$ (equivalently an element $x \in \mathfrak{o}_G$ ).

Definition 5·11. The Moerdijk–Weiss dendroidal category, denoted $\boldsymbol{\Omega}$ , is the full subcategory of $\textbf{U}_{/\mathfrak{o}}$ consisting of those graphs G so that $\operatorname{out}\!(v)$ is a singleton set for every $v\in V_G$ . Alternatively, it consists of those graphs G so that $\operatorname{out} \colon \operatorname{Emb}\!(G) \to \mathbb{N}E_G$ factors through $E_G \hookrightarrow \mathbb{N}E_G$ .

Proof sketch. On objects the value of F is given in the same was as in [ Reference Hackney, Robertson and Yau29 , section 9·2·1]. Briefly, if G is a directed graph then F(G) is the free $E_G$ -colored wheeled properad which has one generator for each vertex $v\in V_G$ . For each vertex v, choose orderings $\operatorname{in}\!(v) = e_1, \dots, e_m$ and $\operatorname{out}\!(v) = e'_{\!\!1}, \dots, e'_{\!\!n}$ for the inputs and output edges of v, and regard v as a generator in the set

(5·1) \begin{equation} v\in F(G)(e_1, \dots, e_m;\; e'_{\!\!1}, \dots, e'_{\!\!n}).\end{equation}

If H is a graph equipped with a total ordering of the sets $\operatorname{in}\!(H) = \{i_1, \dots, i_k\}$ and $\operatorname{out}\!(H) = \{o_1, \dots, o_j\}$ , then an étale map $\mu \colon H \to G$ determines an element in

\begin{align*} F(G)(\mu i_1, \dots, \mu i_k;\; \mu o_1, \dots, \mu o_j).\end{align*}

Given a graphical map $\varphi \colon G \to G'$ , using that F(G) is a free wheeled properad we can specify a map of wheeled properads (see [ Reference Hackney, Robertson and Yau29 , lemma 9·23])

\begin{align*} F\varphi = f \colon F(G) \longrightarrow F(G')\end{align*}

by defining the color map $E_G \to E_{G^{\prime}}$ to be $\bar \varphi_0$ , and value on the generator (5·1),

\begin{align*} f_1(v) \in F(G)(fe_1, \dots, fe_m;\;\, fe'_{\!\!1}, \dots, fe'_{\!\!n}),\end{align*}

is given by the embedding $\varphi_1(v)$ . Composition is defined using graph substitution, and associativity of composition is proved as in [ Reference Hackney, Robertson and Yau34 , proposition 2·25].

Definition 5·16. Let $f \colon G \rightarrowtail H$ be an embedding of (undirected or directed) connected graphs (in $\textbf{U}_{\textrm{int}}$ or $\textbf{U}^{\textrm{int}}_{/\mathfrak{o}}$ ). A graph complement of f consists of a triple $(K,v^G,\alpha)$ where K is another connected graph, $v^G \in V_K$ is a distinguished vertex, and is an active map. This data must satisfy $\alpha_1(v^G) = [f]$ and, for all other vertices $v \in V_K\setminus \{v^G\}$ , there is a vertex $w\in V_H$ with $\alpha_1(v) = [\iota_w]$ .

Proof. Suppose $(K, v^G, \alpha)$ and $(J, u^G, \beta)$ are two graph complements of an embedding $f \colon G \rightarrowtail H$ . The isomorphisms

\begin{align*} V_K \setminus \{v^G \} \xrightarrow{\cong} V_H \setminus f(V_G) \xleftarrow{\cong} V_J \setminus \{u^G \}\end{align*}

extend to a bijection $z_V \colon V_K \to V_J$ sending $v^G$ to $u^G$ .

By Definition 2·9(ii) we have

and these isomorphisms $nb(v) \to nb(z_V(v))$ assemble into a bijection

\begin{align*} z_D \colon D_K = \coprod_{v\in V_K} nb(v) \to \coprod_{u \in V_J} nb(u) = D_J\end{align*}

over $z_V \colon V_K \cong V_J$ . Finally, since $\alpha$ and $\beta$ are active we have isomorphisms

which gives

\begin{align*} z_A \colon A_K = D_K \amalg \eth(K) \xrightarrow{\cong} D_J \amalg \eth(J) = A_J.\end{align*}

If $a \in nb(v)$ (resp. $a \in \eth(K)$ ) then $z_A(a) = z_D(a) \in nb(z_V(v))$ (resp. $z_A(a) \in \eth(J)$ ) is the unique element satisfying $\alpha_0(a) = \beta_0(z_A(a))$ .

The only thing to check is that $z_A\colon A_K \to A_J$ preserves the involution.

If G is not an edge, then $\alpha_0$ and $\beta_0$ are injective by lemma 2·9 of [ Reference Hackney, Robertson and Yau33 ], so $z_A(a)$ is defined by the equation $\alpha_0(a) = \beta_0(z_A(a))$ with no stipulation about where a or $z_A(a)$ live. Using this equation for both a and $a^\dagger$ , we have

\begin{align*} \beta_0(z_A(a^\dagger)) = \alpha_0(a^\dagger) = \alpha_0(a)^\dagger = \beta_0(z_A(a))^\dagger = \beta_0(z_A(a)^\dagger),\end{align*}

hence $z_A(a^\dagger) = z_A(a)^\dagger$ .

If G is an edge, then again using lemma 2·9 of [ Reference Hackney, Robertson and Yau33 ] we have $\alpha_0$ is injective when restricted to $A_K \setminus nb(v^G)$ . Writing $nb(v^G) = \{d_0,d_1\}$ , the only identifications that $\alpha_0$ makes are $\alpha_0(d_0) = \alpha_0(d_1^\dagger)$ and $\alpha_0(d_1) = \alpha_0(d_0^\dagger)$ . A similar situation occurs with $\beta_0$ and $nb(u^G)$ . Thus, as in the previous paragraph, $z_A(a^\dagger) = z_A(a)^\dagger$ so long as $a\notin \{d_0,d_1,d_0^\dagger,d_1^\dagger\}$ . But now $nb(u^G) = \{ z_A(d_0), z_A(d_1) \}$ , and the only way for our bijection to not be involutive would be if $z_A(d_0^\dagger) = z_A(d_1)^\dagger$ and $z_A(d_1^\dagger) = z_A(d_0)^\dagger$ . This cannot happen, as

\begin{align*} \beta_0 (z_A(d_0^\dagger)) = \alpha_0(d_0^\dagger) = \alpha_0(d_1)\neq \alpha_0(d_1)^\dagger= \beta_0(z_A(d_1))^\dagger.\end{align*}

Proof. We can ignore directionality when constructing the graph complement: if the embedding f is a map in $\textbf{U}_{/\mathfrak{o}}$ , then the produced undirected graph K becomes directed via the composite map of presheaves $K \to H \to \mathfrak{o}$ .

We first address the special case when G does not have a vertex. If G is an edge hitting arcs $a_0, a_1 \in A_H$ (with $a_0^\dagger = a_1$ ), we set $V_K = V_H \amalg \{ v^G \}$ , $nb(v^G)$ consists of two new elements $d_0$ and $d_1$ , $D_K = D_H \amalg nb(v^G)$ , $A_K = A_H \amalg \{d_0,d_1\}$ , and all structure the same except that $a_i^\dagger$ is redefined to be $d_i$ for $i=0,1$ . The definition of $\alpha \colon K \to H$ is forced by the requirements in Definition 5·16.

Now suppose that G has at least one vertex. Define the following three sets

\begin{align*}V_K &= (V_H \setminus f(V_G)) \amalg \{v^G\} \cong V_H / {\sim} \\[5pt] D_K &= D_H \setminus \{ f(d) \mid d, d^\dagger \in D_G \} \\[5pt] A_K &= A_H \setminus \{ f(d) \mid d, d^\dagger \in D_G \}.\end{align*}

The quotient presentation for $V_K$ identifies all of the vertices of $f(V_G) \subseteq V_H$ to a single vertex $v^G$ . We thus have a map $D_K \hookrightarrow D_H \to V_H \to V_H/{\sim} \cong V_K$ . The set $A_K$ inherits an involutive structure from $A_H$ . This data defines a graph K. Notice that $\eth(K) = A_K \setminus D_K = A_H \setminus D_H = \eth(H)$ .

There is a clear candidate for a graphical map $\alpha \colon K \to H$ which is an inclusion on arcs and satisfies $\varphi_1(v) = [\iota_v]$ for $v\in V_H \setminus f(V_G)$ and $\varphi_1(v) = [f]$ for $v=v^G$ . This satisfies the conditions from Definition 2·9; the only thing to note is that $nb(v^G) = f(\eth(G)^\dagger)$ which is used in showing (ii). The triple $(K,v^G,\alpha)$ satisfies the conditions of Definition 5·16.

Proof. Suppose $f \colon G \rightarrowtail G'$ is an embedding, which determines a map of wheeled properads $F(f) \colon F(G)\to F(G')$ by Proposition 5·15. Let be the graph complement of f from Lemma 5·19. Since $\alpha$ is active, it induces an isomorphism $K\{G\} \cong G'$ by definition 1·39 and proposition 2·3 of [ Reference Hackney, Robertson and Yau33 ]; here $K\{G\}$ is the graph substitution of G into the vertex $v^G$ of K (see construction 1·18 of [ Reference Hackney, Robertson and Yau33 ]). By theorem 9·52 of [ Reference Hackney, Robertson and Yau29 ], $F(G) \to F(G')$ is a subgraph.

Now suppose that $f \colon F(G) \to F(G')$ is a map of wheeled properads which is a subgraph. By theorem 9·52 of [ Reference Hackney, Robertson and Yau29 ] there is a graph substitution decomposition $K\{G\} \cong G'$ so that f sends the edges and vertices in G to their corresponding images in $K\{G\}$ through this isomorphism. As we always have an embedding $G \rightarrowtail K\{G\}$ , we see that the composition $G \rightarrowtail K\{G\} \cong G'$ is an embedding in $\textbf{U}_{/\mathfrak{o}}$ . The two indicated constructions are inverses to one another.

Proof of Theorem 5·13. Recall from Remark 5·14 that we are taking the objects of the wheeled properadic graphical category, denoted in this proof by $\textbf{C}$ , to be the same as the objects of $\textbf{U}_{/\mathfrak{o}}$ . This category comes with a faithful (but non-full) functor $\textbf{C} \to \textbf{WPpd}$ which has the same action on objects as the functor $F \colon \textbf{U}_{/\mathfrak{o}} \to \textbf{WPpd}$ from Proposition 5·15. Our strategy is to exhibit a lift

(5·2)

which is the identity-on-objects, makes the triangle commute, and is fully-faithful.

Suppose G and G ^′ are objects in $\textbf{U}_{/\mathfrak{o}}$ , and let $f \colon F(G) \to F(G')$ be a map of wheeled properads. The image ([ Reference Hackney, Robertson and Yau29 , definition 9·55]) of f is the G ^′-decorated graph $(f_0G)\{f_1(u)\}$ . Namely, $f_1(u)$ is given by a connected graph $K_u$ together with an étale map $K_u \to G'$ , and $(f_0G)\{f_1(u)\}$ is the (connected) graph substitution $G\{K_u\}_{u\in V_G}$ together with the induced étale map. An embedding $h\colon H \rightarrowtail G$ determines a subgraph $F(H) \to F(G)$ by the equivalence of Lemma 5·20, and there is an associated image of this subgraph given as the étale map

(5·3) \begin{equation} H\{K_{hv}\}_{v\in V_H} \longrightarrow G\{K_{u}\}_{u\in V_G} \longrightarrow G'. \end{equation}

Leaning on the equivalence from Lemma 5·20, by [ Reference Hackney, Robertson and Yau29 , theorem 9·62] the map f constitutes a morphism in $\textbf{C}(G,G')$ if and only if for every embedding h the étale map (5·3) is an embedding. When $f= F\varphi$ , by Construction 4·5 we have that (5·3) is an embedding representing $\hat \varphi [h]$ , hence $F\varphi$ is in $\textbf{C}(G,G')$ . On the other hand, if $f\in \textbf{C}(G,G')$ is an map in the wheeled properadic graphical category, we can define a graphical map $\varphi \colon G \to G'$ by setting $\bar \varphi_0 = f_0 \colon E_G \to E_G$ and letting $\hat \varphi [h]$ be the embedding (5·3) (or using the special case when $h = \iota_v$ and letting $\varphi_1(v)$ be represented by the embedding $K_v \rightarrowtail G'$ associated to $f_1(v)$ ). Using properties of graph substitution, one checks that $(\bar \varphi_0, \hat \varphi)$ satisfies the conditions from Proposition 5·10, hence constitutes a morphism in $\textbf{U}_{/\mathfrak{o}}(G,G')$ .

By looking at their actions on edges and vertices, we see that the two assignments $\varphi \mapsto F\varphi$ and $f \mapsto \varphi$ are inverses. Thus the desired identity-on-objects lift (5·2) of F exists and is fully-faithful.

Proof. It is a classical fact that left Kan extension along a Grothendieck opfibration may be computed by taking colimits over the fibers. The functor $\textbf{C}^{\textrm{op}} \to \textbf{D}^{\textrm{op}}$ is a discrete opfibration by Lemma 6·15, so the fibers $\mathfrak{o}_G$ are discrete, and the result follows.

Proof. For concreteness, we prefer to take a skeleton of $\textbf{D}^\textrm{el}_{/G}$ , with objects

\begin{align*} \{ \iota_v \colon {\unicode{x2606}}_v \rightarrowtail G \mid v\in V_G \} \amalg \{ \kappa_e \colon {{\updownarrow}}_e \rightarrowtail G \mid e\in E_G \}\end{align*}

(see Example 2·5).

We first observe that $\mathfrak{o}$ from Definition 5·3 is Segal. This means that the function

(6·2) \begin{equation} \mathfrak{o}_G \longrightarrow \lim_{k\colon K \rightarrowtail G} \mathfrak{o}_K,\end{equation}

sending $x\in \mathfrak{o}_G$ to the element $(k^*x)$ , is a bijection. It is automatically an injection since the composite $\mathfrak{o}_G \to \lim_k \mathfrak{o}_K \to \prod_{\kappa_e} \mathfrak{o}_{{{\updownarrow}}_e}$ is a bijection, as both $\mathfrak{o}_G$ and $\prod_{\kappa_e} \mathfrak{o}_{{{\updownarrow}}_e}$ are isomorphic to the set of involutive maps $A_G \to \{+1,-1\}$ . On the other hand, if $(y^k) \in \lim_k \mathfrak{o}_K$ , then by using the bijection $\mathfrak{o}_G \cong \prod_{\kappa_e} \mathfrak{o}_{{{\updownarrow}}_e}$ we obtain $x \in \mathfrak{o}_G$ with $\kappa_e^* x = y^{\kappa_e}$ for all $e\in E_G$ . We wish to show that $k^* x = y^k$ for all $k\in \textbf{D}^\textrm{el}_{/G}$ . If G does not have any vertices, then we are done. Otherwise, let v be a vertex. For each $d\in nb(v) = D_{{\unicode{x2606}}_v}$ spanning an edge $e \in E_G$ , there is an embedding $j_d \colon {{\updownarrow}}_e \rightarrowtail {\unicode{x2606}}_v$ sending d to d and making the following triangle commute:

Then $j_d^*(y^{\iota_v}) = y^{\kappa_e} = \kappa_e^* x = j_d^*(\iota_v^*x)$ . Since $\mathfrak{o}_{{\unicode{x2606}}_v} \cong \prod_{nb(v)} \mathfrak{o}_{{{\updownarrow}}_e}$ , we conclude that $y^{\iota_v} = \iota_v^*x$ . We have now established that $y^k = k^* x$ for all $k\in \textbf{D}^\textrm{el}_{/G}$ , proving that (6·2) is a bijection.

Elements on the left-hand side of (6·1) are of the form $(x,(z^k))$ where $x\in \mathfrak{o}_G$ and $z^k \in Z_{(K, k^*x)}$ for each $k\colon K \rightarrowtail G$ in $\textbf{D}^\textrm{el}_{/G}$ . These are subject to the compatibility condition $j^*(z^k) = z^{kj}$ whenever $j \colon K' \rightarrowtail K$ is an embedding. Likewise, elements on the right-hand side of (6·1) are of the form $(y^k,s^k)$ with $y^k \in \mathfrak{o}_K$ , $s^k \in Z_{(K,y^k)}$ for each $k\colon K \rightarrowtail G$ in $\textbf{D}^\textrm{el}_{/G}$ . These are subject to the compatibility conditions $j^*(y^k) = y^{kj}$ and $j^*(s^k) = s^{kj}$ whenever $j \colon K' \rightarrowtail K$ is an embedding. The map (6·1) takes $(x, (z^k))$ to $(k^*x, z^k)$ . Bijectivity of (6·1) follows from that of (6·2).

Proof. Let $Z\in \widehat{\textbf{C}}$ be a Segal object, that is, suppose that

\begin{align*} Z_{(G,x)} \longrightarrow \lim_{(\textbf{C}^\textrm{el}_{/(G,x)})^{\textrm{op}}} Z_{(K,y)}\end{align*}

is a bijection for each $(G,x)\in \textbf{C}$ . We know that $\textbf{C}^\textrm{el}_{/(G,x)} \to \textbf{D}^\textrm{el}_{/G}$ is an equivalence by Lemma 6·16, so this becomes

\begin{align*} Z_{(G,x)} \cong \lim_{k \colon K \rightarrowtail G} Z_{(K,k^*x)}\end{align*}

instead, where K ranges over elementary objects of $\textbf{D}$ . The function

(6·3) \begin{equation} (f_!Z)_G \longrightarrow \lim_{k \colon K \rightarrowtail G} (f_!Z)_K\end{equation}

is such that the following diagram commutes for every $x_0$ in $\mathfrak{o}_G$ and $k_0$ in $\textbf{D}^\textrm{el}_{/G}$ :

Applying Lemma 6·22, we see that (6·3) is a bijection, hence $f_!Z$ is Segal.