Proof. This is just an application of [Reference Fresse17, theorem 12.3.A and proposition 12.3.2]. We just have to check that $\mathcal {O}$ is $\mathbb {S}$-cofibrant as an operad in $\operatorname {bCh}_{\operatorname {CE}}$, i.e. its underlying $\mathbb {S}$-module is cofibrant in $\operatorname {bCh}_{\operatorname {CE}}^{\mathbb {S}}$ with the projective model structure.

Let $S^{0,q}$ be the bicomplex which is just $\Bbbk$ concentrated in bidegree $(0,\,q)$, $q\in \mathbb {Z}$. The maps $0\rightarrowtail S^{0,q}$ are some of the generating cofibrations of $\operatorname {bCh}_{\operatorname {CE}}$. Hence, any $\mathbb {S}$-module $M$ in $\operatorname {bCh}_{\operatorname {CE}}$ concentrated in horizontal degree $0$ with trivial differential and such that each $M(r)$ is $\mathbb {S}_r$-projective, $r\geq 0$, is cofibrant in $\operatorname {bCh}_{\operatorname {CE}}^{\mathbb {S}}$.

Proof. According to [Reference Loday and Vallette28, §8.6.1] we have to check that four diagrams, called (i), (ii), (I) and (II), commute. The commutativity of (i) is equivalent to the first formula in definition 3.5. For (ii) we use that $\varphi (\epsilon ;1)=(1;\epsilon )$ by the second formula.

In order to check (I) we take an element of $\mathcal {D}\circ \mathcal {O}\circ \mathcal {O}$, that we can denote in either way

\[ (\epsilon;(\mu;\nu_{1},\dots, \nu_{r})) = ((\epsilon;\mu);\nu_{1},\dots, \nu_{r}) \]

according to how we associate the triple composite. Here $\nu _i\in \mathcal {O}(s_i)$. On the one hand, if we apply the operad composition $\gamma _{\mathcal {O}}\colon \mathcal {O}\circ \mathcal {O}\to \mathcal {O}$ to the part in $\mathcal {O}\circ \mathcal {O}$ and then $\varphi$ we obtain

\[ \varphi(\epsilon;\mu(\nu_{1},\dots, \nu_{r}) ) = ({-}1)^{|{\mu}|+\sum_{i=1}^{r}|{\nu_i}|}\sum_{i=1}^{\sum_{j=1}^{r}s_j}\mu(\nu_{1},\dots, \nu_{r})(1,\stackrel{i-1}{\dots},1,\epsilon,1,\dots,1). \]

On the other hand,

\begin{align*} & (\varphi(\epsilon;\mu);\nu_{1},\dots, \nu_{r})= ({-}1)^{|{\mu}|}\sum_{i=1}^{r}((\mu;1,\stackrel{i-1}{\dots},1,\epsilon,1,\stackrel{r-i}{\dots},1);\nu_{1},\dots, \nu_{r})\\ & \quad =\sum_{i=1}^{r}({-}1)^{|{\mu}|+\sum_{j=1}^{i-1}|{\nu_j}|}(\mu;(1;\nu_1),\dots,(1;\nu_{i-1}),(\epsilon;\nu_i),(1;\nu_{i+1}),\dots,(1;\nu_{r})). \end{align*}

If we now apply $\varphi$ to all factors in $\mathcal {D}\circ \mathcal {O}$ we obtain

\begin{align*} & \sum_{i=1}^{r}\sum_{j=1}^{s_i}({-}1)^{|{\mu}|+\sum_{j=1}^{i}|{\nu_j}|}(\mu;(\nu_1;1,\dots,1),\dots,(\nu_i;1,\stackrel{j-1}{\dots},\epsilon,\dots,1),\dots,(\nu_{r};1,\dots,1))\\ & \quad= ({-}1)^{|{\mu}|+\sum_{i=1}^{r}|{\nu_i}|}\sum_{i=1}^{\sum_{j=1}^{r}s_j}((\mu;\nu_{1},\dots, \nu_{r});(1,\stackrel{i-1}{\dots},1,\epsilon,1,\dots,1)). \end{align*}

If we now apply $\gamma _{\mathcal {O}}$ we obtain the same result as above.

Finally, for (II) we consider

\[ (\epsilon;(\epsilon;\mu))=((\epsilon;\epsilon);\mu). \]

If we apply the operad composition $\gamma _{\mathcal {D}}$ of $\mathcal {D}$ we obtain $0$ because $\epsilon ^{2}=0$. Moreover,

\begin{align*} (\epsilon;\varphi(\epsilon;\mu)) & =({-}1)^{|{\mu}|}\sum_{i=1}^{r}(\epsilon;(\mu;1,\stackrel{i-1}{\dots},\epsilon,\dots,1)) \\ & =({-}1)^{|{\mu}|}\sum_{i=1}^{r}((\epsilon;\mu);1,\stackrel{i-1}{\dots},\epsilon,\dots,1), \end{align*}

and

\begin{align*} & ({-}1)^{|{\mu}|}\sum_{i=1}^{r}(\varphi(\epsilon;\mu);1,\stackrel{i-1}{\dots},\epsilon,\dots,1) =\sum_{i,j=1}^{r}((\mu;1,\stackrel{j-1}{\dots},\epsilon,\dots,1);1,\stackrel{i-1}{\dots},\epsilon,\dots,1)\\ & \quad= \sum_{1\leq j< i\leq n} (\mu;(1;1),\dots,\underbrace{(\epsilon;1)}_{j \text{ slot}},\dots,\underbrace{(1;\epsilon)}_{i \text{ slot}},\dots,(1;1))\\ & \qquad+ (\mu;(1;1),\dots,(\epsilon;\epsilon),\dots,(1;1))\\ & \qquad- \sum_{1\leq i< j\leq n} (\mu;(1;1),\dots,\underbrace{(1;\epsilon)}_{i \text{ slot}},\dots,\underbrace{(\epsilon;1)}_{j \text{ slot}},\dots,(1;1)). \end{align*}

If we now apply $\gamma _{\mathcal {D}}$ and use that $\epsilon ^{2}=0$ the result is again

\[ \sum_{1\leq j< i\leq n} (\mu; 1,\dots, \overset{j}{\epsilon},\dots, \overset{i}{\epsilon},\dots, 1) -\sum_{1\leq i< j\leq n} (\mu; 1,\dots, \overset{i}{\epsilon},\dots, \overset{j}{\epsilon},\dots, 1)=0. \]

Proof. Let $\mathbb{1}$ be the initial operad in $\operatorname {GrCh}$, which is the monoidal unit for the circle product $\circ$ in $\operatorname {GrCh}^{\mathbb {S}}$, and let $\eta _{\mathcal {P}}\colon \mathbb{1}\to \mathcal {P}$ be the unit of a generic operad $\mathcal {P}$ in $\operatorname {GrCh}$.

We have operad morphisms induced by the units of $\mathcal {O}$ and $\mathcal {D}$, respectively,

Therefore, any $d{\mathcal {O}}$-algebra $A$ is an $\mathcal {O}$-algebra in $\operatorname {GrCh}$ as well as a bicomplex, see remark 3.4.

The second equation of definition 3.5 implies

\[ d_h(\mu(x_1,\dots,x_r))=\sum_{s=1}^{r}({-}1)^{|{\mu}|+\sum_{t=1}^{s-1}|{x_t}|}\mu(x_1,\dots, d_h(x_s),\dots,x_r), \]

This shows that $A$ is not only an $\mathcal {O}$-algebra as a graded complex but as an unbounded bicomplex. The converse is essentially the same.

Proof. Let $\mathcal {P}$ be the operad defined by the presentation in the statement. On the one hand, there is an operad morphism $\mathcal {P}\to d{\mathcal {O}}$ defined on generators by the inclusions

\[ E\cong E\circ\mathbb{1}\longrightarrow \mathcal{O}\circ_{\varphi}\mathcal{D}\longleftarrow\mathbb{1}\circ\Bbbk\cdot\epsilon\cong\Bbbk\cdot\epsilon. \]

These arrows are given by the inclusions of generators $E\subset \mathcal {O}$ and $\Bbbk \cdot \epsilon \subset \mathcal {D}$ and the corresponding operadic units.

On the other hand, it is easy to check that

\[ (\mathcal{O}\circ\mathcal{D})(r)=\mathcal{O}(r)\otimes\mathcal{D}^{{\otimes}^{r}} \]

as $\mathbb {S}_r$-modules, with the permutation action on the tensor power and the diagonal action on the tensor product. Hence, there is an $\mathbb {S}$-module morphism

\begin{align*} \mathcal{O}\circ\mathcal{D} & \longrightarrow\mathcal{P}, \\ (\mu;\epsilon^{i_1},\dots,\epsilon^{i_r}) & \;\mapsto\;\mu(\epsilon^{i_1},\dots,\epsilon^{i_r}). \end{align*}

Here, each $i_j\in \{0,\,1\}$. This morphism is surjective because of the relations in $D$.

The composite

\[ \mathcal{O}\circ\mathcal{D}\longrightarrow\mathcal{P}\longrightarrow d{\mathcal{O}} \]

is clearly the identity on the underlying $\mathbb {S}$-module $\mathcal {O}\circ \mathcal {D}$. Since the first one is surjective, both are isomorphisms.

Proof. The ‘only if’ part follows by definition. If we now assume the existence of $\mathcal {C}$ with the required properties then there is an adjoint cooperad morphism $\bar {\zeta }\colon \mathcal {C}\to B\mathcal {O}$ [Reference Loday and Vallette28, theorem 6.5.7]. Here $B\mathcal {O}$ is the bar construction, see [Reference Fresse16, §3.1.9] and [Reference Loday and Vallette28, §6.5.1]. The morphism $\bar {\zeta }$ is a quasi-isomorphism by the left module version of [Reference Loday and Vallette28, theorem 2.1.13] and [Reference Loday and Vallette28, lemma 4.7.2], see also [Reference Loday and Vallette28, theorem 6.6.1]. Since $\mathcal {C}$ has trivial differential (it is a cooperad in $\operatorname {GrMod}$), it is the homology of the bar construction $B\mathcal {O}$. Moreover, since $\zeta$ vanishes in weights $\neq 1$, $\bar {\zeta }$ maps to the maximal weight part of $B\mathcal {O}$ on each bar degree, see [Reference Fresse16, lemma 5.2.2]. Therefore, the lemma follows, see [Reference Fresse16, §5.2.3].

Proof. Let $\kappa \colon \Bbbk [\delta ]\to \Bbbk [\epsilon ]/(\epsilon ^{2})$ be the morphism of vertical degree $-1$ defined by $\kappa (\delta )=\epsilon$ and $\kappa (\delta ^{r})=0$ for $r\neq 1$. This is a twisting morphism. The differential of the corresponding Koszul complex $\Bbbk [\epsilon ]/(\epsilon ^{2})\otimes _{\kappa }\Bbbk [\delta ]$ is $d(1\otimes \delta ^{r})=\epsilon \otimes \delta ^{r-1}$ for $r>0$, $d(1\otimes 1)=0$, and $d(\epsilon \otimes \delta ^{r})=0$ for $r\geq 0$. This is clearly acyclic.

Proof. We have to check that the categorical duals of the four diagrams in [Reference Loday and Vallette28, §8.6.1] commute. These diagrams are called (i), (ii), (I) and (II). By the very definition,

\[ {\varphi}^{\unicode{x00A1}}(1;\mu)=(\mu;1,\dots,1),\quad {\varphi}^{\unicode{x00A1}}(\delta^{i};1)=(1;\delta^{i}). \]

We choose an element $(\delta ^{i};\mu )\in ({\mathcal {D}}^{\unicode{x00A1}}\circ {\mathcal {O}}^{\unicode{x00A1}})(r)$ and decompose the ${\mathcal {O}}^{\unicode{x00A1}}$ term, using the Sweedler notation in remark 5.23,

\begin{align*} (\delta^{i},\Delta(\mu)) & =\sum_{[\mu]}(\delta^{i};(\nu;\nu^{1},\dots,\nu^{l})\cdot\tau) \\ & =\sum_{[\mu]}((\delta^{i};\nu);\nu^{1},\dots,\nu^{l})\cdot\tau. \end{align*}

We apply ${\varphi }^{\unicode{x00A1}}$ to the ${\mathcal {D}}^{\unicode{x00A1}}\circ {\mathcal {O}}^{\unicode{x00A1}}$ component

\begin{align*} \sum_{[\mu]}({\varphi}^{\unicode{x00A1}}(\delta^{i};\nu);\nu^{1},\dots,\nu^{l})\cdot\tau & = \sum_{[\mu]}\sum_{j_1+\cdots+j_l=i}((\nu;\delta^{j_1},\dots,\delta^{j_l});\nu^{1},\dots,\nu^{l})\cdot\tau \\ & =\sum_{[\mu]}\sum_{j_1+\cdots+j_l=i}(\nu;(\delta^{j_1};\nu^{1}),\dots,(\delta^{j_l};\nu^{l}))\cdot\tau. \end{align*}

We now apply ${\varphi }^{\unicode{x00A1}}$ to all factors in ${\mathcal {D}}^{\unicode{x00A1}}\circ {\mathcal {O}}^{\unicode{x00A1}}$. If $\nu ^{j}\in {\mathcal {O}}^{\unicode{x00A1}}(s_j)$, this yields

\begin{align*} & \sum_{[\mu]}\sum_{j_1+\cdots+j_l=i}\big(\nu;\sum_{k_{1,1}+\dots+k_{1,s_1}=j_1}(\nu^{1};\delta^{k_{1,1}},\dots,\delta^{k_{1,s_1}}),\dots,\\ & \quad \sum_{k_{l,1}+\dots+k_{l,s_l}=j_l}(\nu^{l};\delta^{k_{l,1}},\dots,\delta^{k_{l,s_l}})\big)\cdot\tau. \end{align*}

The composite $\mathcal {C}\circ {\mathcal {D}}^{\unicode{x00A1}}$ is given by

\[ (\mathcal{C}\circ{\mathcal{D}}^{\unicode{x00A1}})(r)=\mathcal{C}(r)\otimes({\mathcal{D}}^{\unicode{x00A1}})^{{\otimes}^{r}} \]

as $\mathbb {S}_r$-modules, with the permutation action on the tensor power and the diagonal action on the tensor product, like in the proof of proposition 3.8.

Since necessarily $\sum _{j=1}^{l}s_j=r$, the previous summation coincides with

\begin{align*} & \sum_{[\mu]}\sum_{j_1+\cdots+j_r=i}((\nu;\nu^{1},\dots,\nu^{l});\delta^{j_1},\dots,\delta^{j_r})\cdot\tau\\ & \quad= \sum_{j_1+\cdots+j_r=i}\sum_{[\mu]}((\nu;\nu^{1},\dots,\nu^{l})\cdot\tau;\delta^{j_{\tau(1)}},\dots,\delta^{j_{\tau(r)}}). \end{align*}

Changing variables, this equals

\[ \sum_{j_1+\cdots+j_r=i}\sum_{[\mu]}((\nu;\nu^{1},\dots,\nu^{l})\cdot\tau;\delta^{j_{1}},\dots,\delta^{j_{r}}). \]

This is the same as if we first consider ${\varphi }^{\unicode{x00A1}}(\delta ^{i},\,\mu )$ in definition 4.5 and then decompose $\mu \in {\mathcal {O}}^{\unicode{x00A1}}(r)$. This proves the dual of (II).

We can also start by applying the decomposition to the ${\mathcal {D}}^{\unicode{x00A1}}$ term $\delta ^{i}$,

\begin{align*} (\Delta(\delta^{i}),\mu) & =\sum_{j+k=i}((\delta^{j};\delta^{k});\mu) \\ & =\sum_{j+k=i}(\delta^{j};(\delta^{k};\mu)), \end{align*}

then we apply $\varphi$,

\begin{align*} \sum_{j+k=i}(\delta^{j};\varphi(\delta^{k};\mu)) & = \sum_{j+k=i}\sum_{l_1+\dots+l_r=k}(\delta^{j};(\mu;\delta^{l_1},\dots,\delta^{l_r})) \\ & =\sum_{j+k=i}\sum_{l_1+\dots+l_r=k}((\delta^{j};\mu);\delta^{l_1},\dots,\delta^{l_r})), \end{align*}

and $\varphi$ again,

\begin{align*} & \sum_{j+k=i}\sum_{l_1+\dots+l_r=k}(\varphi(\delta^{j};\mu);\delta^{l_1},\dots,\delta^{l_r})\\ & \quad= \sum_{j+k=i}\sum_{l_1+\dots+l_r=k}\sum_{m_1+\dots+m_r=j}((\mu;\delta^{m_1},\dots,\delta^{m_r});\delta^{l_1},\dots,\delta^{l_r})\\ & \quad= \sum_{j+k=i}\sum_{l_1+\dots+l_r=k}\sum_{m_1+\dots+m_r=j}(\mu;(\delta^{m_1};\delta^{l_1}),\dots,(\delta^{m_r};\delta^{l_r}))\\ & \quad= \sum_{l_1+\dots+l_r+m_1+\dots+m_r=i}(\mu;(\delta^{m_1};\delta^{l_1}),\dots,(\delta^{m_r};\delta^{l_r})). \end{align*}

This is the same as if we take $\varphi (\delta ^{i};\mu )$ in definition 4.5 and decompose all terms in ${\mathcal {D}}^{\unicode{x00A1}}$,

\begin{align*} & \sum_{j_1+\cdots+j_r=i}\big(\mu;\sum_{m_1+l_1=j_1}(\delta^{m_1};\delta^{l_1}),\dots,\sum_{m_r+l_r=j_r}(\delta^{m_r};\delta^{l_r})\big)\\ & \quad= \sum_{m_1+l_1+\cdots+m_r+l_r=i}(\mu;(\delta^{m_1};\delta^{l_1}),\dots,(\delta^{m_r};\delta^{l_r})). \end{align*}

This checks the dual of (I).

In order to show the commutativity of the dual of (i), we take $\varphi (\delta ^{i};\mu )$ in definition 4.5. Then we apply the counit $\varepsilon _{{\mathcal {D}}^{\unicode{x00A1}}}\colon {\mathcal {D}}^{\unicode{x00A1}}\to \mathbb{1}$ to the ${\mathcal {D}}^{\unicode{x00A1}}$ factors, which yields $(\mu ;1,\,\dots,\,1)$ if $i=0$ and $0$ otherwise. We finally apply the unit isomorphism ${\mathcal {O}}^{\unicode{x00A1}}\circ \mathbb{1}\cong {\mathcal {O}}^{\unicode{x00A1}}$, which yields $\mu$ in the first case. The result is the same as if we first apply the counit $\varepsilon _{{\mathcal {D}}^{\unicode{x00A1}}}$ to the ${\mathcal {D}}^{\unicode{x00A1}}$ factor of $(\delta ^{i};\mu )$, which yields $(1;\mu )$ if $i=0$ and $0$ otherwise, and then the unit isomorphism $\mathbb{1}\circ {\mathcal {O}}^{\unicode{x00A1}}\cong {\mathcal {O}}^{\unicode{x00A1}}$.

The commutativity of the dual of (ii) is easy because the only ${\mathcal {O}}^{\unicode{x00A1}}$ term appearing in $\varphi (\delta ^{i};\mu )$ is $\mu$ itself, see definition 4.5. Therefore, on the one hand, if $\mu$ has positive weight, we obtain zero either if we apply $\varepsilon _{{\mathcal {O}}^{\unicode{x00A1}}}$ to all $\mu$'s in $\varphi (\delta ^{i};\mu )$ or if we directly apply it to the only $\mu$ in $(\delta ^{i};\mu )$. On the other hand, if $\mu =1$ is the operadic unit, then $\varepsilon _{{\mathcal {O}}^{\unicode{x00A1}}}(1)=1$, $\varphi (\delta ^{i};1)=(1,\,\delta ^{i})$, and both $(\delta ^{i};1)$ and $(1,\,\delta ^{i})$ map to $\delta ^{i}$ via the unit isomorphisms ${\mathcal {D}}^{\unicode{x00A1}}\circ \mathbb{1}\cong {\mathcal {D}}^{\unicode{x00A1}}\cong \mathbb{1}\circ {\mathcal {D}}^{\unicode{x00A1}}$.

Proof. Let $\zeta \colon {\mathcal {D}}^{\unicode{x00A1}}\circ _{{\varphi }^{\unicode{x00A1}}}{\mathcal {O}}^{\unicode{x00A1}}\to \mathcal {O}\circ _{\varphi }\mathcal {D}=d{\mathcal {O}}$ be the composite

\[ {\mathcal{D}}^{\unicode{x00A1}}\circ_{{\theta}^{\unicode{x00A1}}}{\mathcal{O}}^{\unicode{x00A1}}\twoheadrightarrow \mathbb{1}\circ sE\oplus\Bbbk\cdot\delta\circ\mathbb{1}\cong sE\oplus\Bbbk\cdot\delta\stackrel{s^{{-}1}}{\longrightarrow} E\oplus\Bbbk\cdot\epsilon\cong E\circ\mathbb{1}\oplus\mathbb{1}\circ\Bbbk\cdot\epsilon\hookrightarrow \mathcal{O}\circ_{\theta}\mathcal{D}. \]

These arrows are the projection onto the weight $1$ part, the desuspension, and the inclusion of the weight $1$ part, respectively, i.e. $\zeta$ vanishes in weight $\neq 1$, and in weight $1$ it is given by

\[ \zeta(\delta;1)= (1;\epsilon),\quad \zeta(1;s \mu)\ = (\mu;1,\dots,1),\ \mu\in E. \]

This degree $-1$ morphism is a twisting morphism, i.e. the composite $\zeta \star \zeta$

vanishes. Let us check this claim.

The infinitesimal decomposition $\Delta _{(1)}$ of a cooperad $\mathcal {C}$ will be denoted by

(4.11)\begin{equation} \Delta_{(1)}\colon\mathcal{C}\longrightarrow \mathcal{C}\circ_{(1)}\mathcal{C},\quad \Delta_{(1)}(\mu)=\sum_{(\mu)}(\mu^{(1)}\circ_{l}\mu^{(2)})\cdot\sigma, \end{equation}

like in [Reference Loday and Vallette28, §10.1.2]. See also (4.1).

With this notation, the infinitesimal decomposition of ${\mathcal {D}}^{\unicode{x00A1}}\circ _{{\varphi }^{\unicode{x00A1}}}{\mathcal {O}}^{\unicode{x00A1}}$ is

\[ \Delta_{(1)}(\delta^{i};\mu)=\sum_{j+k=i}\sum_{(\mu)}((\delta^{j};\mu^{(1)})\circ_l(\delta^{k};\mu^{(2)}))\cdot\sigma,\qquad\mu\in{\mathcal{O}}^{\unicode{x00A1}}. \]

Therefore, $\zeta \star \zeta$ vanishes except possibly in weight $2$. In weight $2$ it also vanishes by the following formulas,

\[ \zeta\star\zeta(\delta^{2};1)= \zeta(\delta;1)\circ_1 \zeta(\delta;1)=(1;\epsilon)\circ_1(1;\epsilon)=(1;\epsilon^{2})=0. \]

Given $\mu \in E$,

\begin{align*} \zeta\star\zeta(\delta;s\mu) & = \zeta(\delta;1)\circ_{1}\zeta(1;s\mu)-({-}1)^{|{\mu}|}\sum_{i=1}^{r} \zeta(1;s\mu)\circ_{i}\zeta(\delta;1) \\ & = (1;\epsilon)\circ_{1}(\mu;1,\dots,1)-({-}1)^{|{\mu}|}\sum_{i=1}^{r} (\mu;1,\dots,1)\circ_{i}(1;\epsilon) \\ & = \varphi(\epsilon;\mu)-({-}1)^{|{\mu}|}\sum_{i=1}^{r} (\mu;1,\stackrel{i-1}{\dots},1,\epsilon,1,\stackrel{r-i}{\dots},1)=0 \end{align*}

by definition of $\varphi$. We have that $({\mathcal {O}}^{\unicode{x00A1}})^{(2)}=s^{2}R$, and given $\sum \mu \circ _l\nu \in R\subset E\circ _{(1)}E$,

\begin{align*} \zeta\star\zeta\left(1;\sum ({-}1)^{|{\mu}|}s\mu\circ_ls\nu\right) & ={-}\sum \zeta(1;s\mu)\circ_l\zeta(1;s\nu) \\ & ={-}\sum (\mu;1,\dots,1)\circ_l(\nu;1,\dots,1) \\ & = \left(-\sum \mu\circ_l\nu;1,\dots,1\right)=0 \end{align*}

by the relations $R$ in $\mathcal {O}$.

We now show that the twisted composite product $(\mathcal {O}\circ _{\varphi }\mathcal {D})\circ _{\zeta }({\mathcal {D}}^{\unicode{x00A1}}\circ _{{\varphi }^{\unicode{x00A1}}}{\mathcal {O}}^{\unicode{x00A1}})$ is acyclic in positive weights. The decomposition of ${\mathcal {D}}^{\unicode{x00A1}}\circ _{{\varphi }^{\unicode{x00A1}}}{\mathcal {O}}^{\unicode{x00A1}}$, which plays a role in the definition of the previous twisted complex, is

\[ \Delta(\delta^{i};\mu)=\sum_{[\mu]}\sum_{j+k_1+\cdots+k_l=i}((\delta^{j};\nu);(\delta^{k_1};\nu^{1}),\dots,(\delta^{k_l};\nu^{l}))\cdot\tau. \]

Here we use the Sweedler formula for the decomposition of ${\mathcal {O}}^{\unicode{x00A1}}$ in (4.1). We increasingly filter $(\mathcal {O}\circ _{\varphi }\mathcal {D})\circ _{\zeta }({\mathcal {D}}^{\unicode{x00A1}}\circ _{{\varphi }^{\unicode{x00A1}}}{\mathcal {O}}^{\unicode{x00A1}})$ according to the weight of ${\mathcal {O}}^{\unicode{x00A1}}$. This is indeed a filtration because, according to the previous formulas

\begin{align*} (\zeta\circ\operatorname{id}_{})\Delta(\delta^{i};\mu)& =((1;\epsilon);(\delta^{i-1};\mu)) \\ & \quad+\sum_{\substack{[\mu]\\\nu\in sE}}\sum_{k_1+\cdots+k_l=i}(s^{{-}1}\nu;1,\dots,1);(\delta^{k_1};\nu^{1}),\dots,(\delta^{k_l};\nu^{l}))\cdot\tau, \end{align*}

and the sum of the weights of the $\nu ^{i}$ is one less than the weight of $\mu$. This filtration is bounded below and exhaustive. Moreover, the previous formula also shows that $E^{0}((\mathcal {O}\circ _{\varphi }\mathcal {D})\circ _{\zeta }({\mathcal {D}}^{\unicode{x00A1}}\circ _{{\varphi }^{\unicode{x00A1}}}{\mathcal {O}}^{\unicode{x00A1}}))=\mathcal {O}\circ (\mathcal {D}\circ _{\kappa }{\mathcal {D}}^{\unicode{x00A1}})\circ {\mathcal {O}}^{\unicode{x00A1}}$, with $d_0$ given just by the Koszul complex differential of $\mathcal {D}$. We have that

\[ \mathcal{O}\circ\mathcal{D}\circ{\mathcal{D}}^{\unicode{x00A1}}\circ{\mathcal{O}}^{\unicode{x00A1}} = \bigoplus_{r\geq 0} \mathcal{O}(r)\otimes_{\mathbb{S}_r}(\mathcal{D}\circ{\mathcal{D}}^{\unicode{x00A1}}\circ{\mathcal{O}}^{\unicode{x00A1}})^{{\otimes}^{r}}. \]

The homology of $\mathcal {D}\circ _{\kappa }{\mathcal {D}}^{\unicode{x00A1}}$ is $\mathbb{1}$ by proposition 4.4. Therefore, the homology of $\mathcal {D}\circ {\mathcal {D}}^{\unicode{x00A1}}\circ {\mathcal {O}}^{\unicode{x00A1}}$ is ${\mathcal {O}}^{\unicode{x00A1}}$. Since each $\mathcal {O}(r)$ is projective as an $\mathbb {S}_r$-module, we conclude that the homology of $\mathcal {O}\circ (\mathcal {D}\circ _{\kappa }{\mathcal {D}}^{\unicode{x00A1}})\circ {\mathcal {O}}^{\unicode{x00A1}}$ is $\mathcal {O}\circ {\mathcal {O}}^{\unicode{x00A1}}$. Moreover, as a graded complex $E^{1}((\mathcal {O}\circ _{\varphi }\mathcal {D})\circ _{\zeta }({\mathcal {D}}^{\unicode{x00A1}}\circ _{{\varphi }^{\unicode{x00A1}}}{\mathcal {O}}^{\unicode{x00A1}}))=\mathcal {O}\circ _{\kappa }{\mathcal {O}}^{\unicode{x00A1}}$ is the Koszul complex of $\mathcal {O}$. Indeed, the spectral sequence differential $d_1$ is the Koszul complex differential because $\mathcal {O}$ is a sub-operad of $d{\mathcal {O}}=\mathcal {O}\circ _{\varphi }\mathcal {D}$, ${\mathcal {O}}^{\unicode{x00A1}}$ is a sub-cooperad of ${\mathcal {D}}^{\unicode{x00A1}}\circ _{{\varphi }^{\unicode{x00A1}}}{\mathcal {O}}^{\unicode{x00A1}}$, and the twisting morphism $\zeta$ restricts to the canonical twisting morphism $\kappa \colon {\mathcal {O}}^{\unicode{x00A1}}\to \mathcal {O}$ on ${\mathcal {O}}^{\unicode{x00A1}}$. Since $\mathcal {O}$ is Koszul, $E^{1}$ is acyclic in positive weights, hence so is $(\mathcal {O}\circ _{\varphi }\mathcal {D})\circ _{\zeta }({\mathcal {D}}^{\unicode{x00A1}}\circ _{{\varphi }^{\unicode{x00A1}}}{\mathcal {O}}^{\unicode{x00A1}})$, see lemma 4.9. Now this theorem follows from lemma 4.3.

Proof. By proposition 4.4, ${\mathcal {D}}^{\unicode{x00A1}}=\Bbbk [\delta ]$ is the polynomial coalgebra on one generator of bidegree $(-1,\,1)$ with trivial differential. Its infinitesimal decomposition is

\[ \Delta_{(1)}(\delta^{i})=\sum_{j+k=i}\delta^{j}\circ_1\delta^{k}. \]

Applying the graded complex version of remark 5.2, the correspondence between twisted complexes and ${\mathcal {D}}_{\infty }$-algebras is given by the equations

\[ d_0(x) =d(x),\quad d_i(x) =\delta^{i}(x),\quad i>0. \]

Equation (5.7) corresponds to (5.4) for ${\mathcal {D}}^{\unicode{x00A1}}$ with the previous infinitesimal decomposition.

Proof. We can split the Sweedler notation for the infinitesimal decomposition of $\mathcal {O}$ in (4.11) as follows. If $\mu \in {\mathcal {O}}^{\unicode{x00A1}}$ has positive weight,

\[ \Delta_{(1)}(\mu)=1\circ_i\mu+\sum_{(\mu)'}(\mu^{(1)}\circ_{l}\mu^{(2)})\cdot\sigma+\sum_{s=1}^{r}\mu\circ_s1, \]

where $(\mu )'$ stands for the summands where both $\mu ^{(1)}$ and $\mu ^{(2)}$ have positive weight, and

\[ \Delta_{(1)}(1)=1\circ_11. \]

The infinitesimal decomposition of ${d{\mathcal {O}}}^{\unicode{x00A1}}$ is then

\begin{align*} \Delta_{(1)}(\delta^{i};\mu) & = \sum_{j+k=i}(\delta^{j};1)\circ_1(\delta^{k};\mu) +\sum_{s=1}^{r}\sum_{j+k=i}(\delta^{j};\mu)\circ_s(\delta^{k};1) \\ & \quad+ \sum_{(\mu)'}\sum_{j+k=i}((\delta^{j};\mu^{(1)})\circ_l(\delta^{k};\mu^{(2)}))\cdot\sigma, \end{align*}

see theorem 4.10.

By the graded version of remark 5.2, a ${d{\mathcal {O}}}_{\infty }$-algebra is a graded complex $A$ with differential $d_0$ equipped with module morphisms

\[ ({\mathcal{D}}^{\unicode{x00A1}}\circ_{{\varphi}^{\unicode{x00A1}}}{\mathcal{O}}^{\unicode{x00A1}})(r)_{p_0,q_0}\otimes A_{p_1,q_1}\otimes\cdots\otimes A_{p_r,q_r} \longrightarrow A_{p_0+p_1+\cdots+p_r,q_0-1+q_1+\cdots+q_r}, \]

satisfying certain equations. Note that ${\mathcal {D}}^{\unicode{x00A1}}\circ {\mathcal {O}}^{\unicode{x00A1}}=k[\delta ]\otimes {\mathcal {O}}^{\unicode{x00A1}}$. If we denote

\begin{align*} (\delta^{i}\otimes\mu)(x_1,\cdots, x_r) & =\mu_i(x_1,\dots,x_r),\quad \mu\text{ of positive weight and }i\geq 0; \\ (\delta^{i}\otimes 1)(x) & =d_i(x),\quad i>0; \\ (\delta^{0}\otimes 1)(x) & =1_0(x); \end{align*}

the equations are

\[ (\mu_i\cdot\sigma)(x_1,\dots,x_r)=({-}1)^{\alpha_\sigma}\mu_i(x_{\sigma^{{-}1}(1)},\dots,x_{\sigma^{{-}1}(r)}) \]

for any permutation $\sigma \in \mathbb {S}_r$, which coincides with (5.11),

(5.14)\begin{align} & d_0(\mu_i(x_1,\dots,x_r))+\sum_{t=1}^{r}({-}1)^{\beta}\mu_i(x_1,\dots,d_0(x_t),\dots,x_r) \\ & \quad+ \sum_{\substack{j+k=i\\ j>0}}d_j\mu_k(x_1,\dots, x_r) +\sum_{s=1}^{r}({-}1)^{\beta}\sum_{\substack{j+k=i\\ k>0}}\mu_j(x_1,\dots,d_k(x_s),\dots,x_r)\nonumber\\ & \quad+ \sum_{(\mu)'}({-}1)^{\gamma}\sum_{j+k=i}\mu_j^{(1)}(x_{\sigma^{{-}1}(1)},\dots,\mu_k^{(2)}(x_{\sigma^{{-}1}(l)},\dots),\dots)=0, \nonumber \end{align}

if $\mu$ has positive weight,

(5.15)\begin{equation} d_0(d_i(x_1))+d_i(d_0(x_1))+\sum_{\substack{j+k=i\\j,k>0}}d_j(d_k(x_1))=0, \end{equation}

and

(5.16)\begin{equation} 1_0(x)=0. \end{equation}

Observe that (5.15) is the equation (5.7) defining twisted complexes. Moreover, (5.14) is (5.12) replacing the summation index $(\mu )$ with $(\mu)'$. We want to avoid this replacement, because $(\mu)'$ has been solely introduced for this proof. For that reason, we are going to make a new convention which will make the summands in $(\mu )$ but not in $(\mu)'$ vanish. This convention is

\[ 1_i(x) = 0,\quad i>0; \]

which, together with (5.16), amounts to (5.13) .

Proof. Since $A$ is filtered, the restriction of the differential to $A_{p,q}$ is determined by its components,

\[ d_i\colon A_{p,q}\longrightarrow A_{p-i,q+i-1},\quad i\geq 0. \]

There are finitely many for each $p$ since $A_{p-i,q+i-1}=0$ for $i>p$. It is well known that (5.7) is equivalent to the differential of $A$ squaring to zero.

Similarly, the restriction to ${\mathcal {O}}^{\unicode{x00A1}}(r)_s\otimes A_{p_1,q_1}\otimes \cdots \otimes A_{p_r,q_r}$ of the structure maps in remark 5.2 are determined by their components

\[ {\mathcal{O}}^{\unicode{x00A1}}(r)_s\otimes A_{p_1,q_1}\otimes\cdots\otimes A_{p_r,q_r} \longrightarrow A_{p_1+\cdots+p_r-i,s-1+q_1+\cdots+q_r+i},\qquad i\geq 0, \]

since these structure maps are compatible with the filtration of $A$. If we denote these components by

\[ \mu\otimes x_1\otimes\cdots\otimes x_r\mapsto \mu_i(x_1,\dots,x_r), \]

the equations in proposition 5.10 are a mere translation of those in remark 5.2.

Proof. Any cofibrant object $X$ in $\operatorname {Ch}$ (endowed with the projective model structure) with degree-wise projective homology $H_*(X)$ is chain homotopy equivalent to $H_*(X)$ endowed with the trivial differential. Indeed, $H_*(X)$ is also cofibrant in $\operatorname {Ch}$ under these assumptions and all objects in $\operatorname {Ch}$ are fibrant. Therefore, any quasi-isomorphism $H_*(X)\to X$ is a homotopy equivalence. Such a quasi-isomorphism (a cycle selection map) can be constructed because $H_*(X)$ is projective. Moreover, any homotopy equivalence $H_*(X)\to X$ is the inclusion of a deformation retract because chain homotopic endomorphisms of $H_*(X)$ are equal since $H_*(X)$ has trivial differential. We can actually make it part of a contraction in the sense of [Reference Berglund5, definition 2.1 and remark 2.1]. This obviously extends to the category of graded complexes $\operatorname {GrCh}=\operatorname {Ch}^{\mathbb {Z}}$ with the product model structure.

Let ${A}_Q\to A$ be a cofibrant resolution in the semi-model category $\operatorname {Alg}_{\operatorname {bCh}_{\operatorname {CE}}}({\mathcal {O}})$ of proposition 2.4. The bicomplex underlying ${A}_Q$ is a Cartan–Eilenberg resolution, see remark 2.5, so ${A}_Q$ has projective vertical homology. We simply denote the vertical homology by $H_*({A}_Q)$, since it is the homology of the underlying graded complex.

The graded complex underlying ${A}_Q$ is cofibrant because the graded complex morphisms underlying the generating cofibrations in $\operatorname {bCh}_{\operatorname {CE}}$ are cofibrations in $\operatorname {GrCh}$, see [Reference Muro and Roitzheim32, §4], and colimits are computed point-wise both in $\operatorname {bCh}$ and $\operatorname {GrCh}$. Therefore, we can take a map $i\colon H_*({A}_Q)\to {A}_Q$ in $\operatorname {GrCh}$ which is part of a contraction. We can now invoke the homotopy transfer theorem, see [Reference Loday and Vallette28, theorem 10.3.1] and [Reference Berglund5, theorem 1.3]. This theorem endows $H_*({A}_Q)$ with a minimal derived homotopy $\mathcal {O}$-algebra structure ${A}_M$ and enhances the map of graded complexes $i\colon H_*({A}_Q)\to {A}_Q$ to a derived $\infty$-morphism $i\colon {A}_M\leadsto {A}_Q$. This derived $\infty$-morphism is an $E^{2}$-equivalence. It actually induces an isomorphism on the $E^{1}$ page of the corresponding spectral sequences, since this term is the vertical homology. If we compose it with ${A}_Q\to A$ we obtain the desired minimal model ${A}_M\leadsto A$.

Proof. First of all, notice that any object in $\operatorname {GrCh}^{\mathbb {S}}$ with trivial differential which is arity-wise projective as a module over the corresponding symmetric group is cofibrant in $\operatorname {GrCh}^{\mathbb {S}}$. We will show that both $d{\mathcal {O}}$ and ${d{\mathcal {O}}}^{\unicode{x00A1}}$ satisfy this property. They clearly have trivial differential. The second one is arity-wise ${d{\mathcal {O}}}^{\unicode{x00A1}}(r)=({\mathcal {D}}^{\unicode{x00A1}}\circ {\mathcal {O}}^{\unicode{x00A1}})(r)=\Bbbk [\delta ]\otimes {\mathcal {O}}^{\unicode{x00A1}}(r)$ by theorem 4.10. As an $\mathbb {S}_r$-module this is just an infinite direct sum of copies of ${\mathcal {O}}^{\unicode{x00A1}}(r)$, which is $\mathbb {S}_r$-projective by the standing hypotheses of this section.

The case of $d{\mathcal {O}}$ is slightly more difficult. We have seen in the proof of proposition 3.8 that

\[ (\mathcal{O}\circ\mathcal{D})(r)=\mathcal{O}(r)\otimes\mathcal{D}^{{\otimes}^{r}} \]

with $\mathbb {S}_r$ acting on the tensor power by permutation of tensor factors and diagonally on the first tensor product. We want to prove that this $\mathbb {S}_r$-module is projective using the fact that $\mathcal {O}(r)$ is. The tensor power is $\Bbbk$-projective. Hence, it suffices to show that, given a group $G$ and a $G$-module $M$, the tensor product $\Bbbk [G]\otimes M$ with the diagonal action is isomorphic as a $G$-module to $M\otimes \Bbbk [G]$ with $G$ acting just on the right tensor factor. It is easy to check that the map

\begin{align*} k[G]\otimes M & \longrightarrow M\otimes k[G], \\ g\otimes x & \;\mapsto\; xg^{{-}1}\otimes g, \end{align*}

provides the desired isomorphism.

Proof. The argument is like in [Reference Loday and Vallette28, theorem 11.4.4], but here we should invoke [Reference Fresse17, proposition 11.5.3 (b)] instead of [Reference Loday and Vallette28, proposition 6.2.3]. We apply [Reference Fresse17, proposition 11.5.3 (b)] to $A$, which is cofibrant in $\operatorname {GrCh}$ by hypothesis, and to the inclusion of the weight $0$ part $\mathbb{1}$ into the Koszul complex $d{\mathcal {O}}\circ _{\zeta }{d{\mathcal {O}}}^{\unicode{x00A1}}$, which is a weak equivalence in $\operatorname {GrCh}^{\mathbb {S}}$ by theorem 4.10. The object $\mathbb{1}$ is obviously cofibrant in $\operatorname {GrCh}^{\mathbb {S}}$. We have to check that the Koszul complex is too. For this, we need to use left modules over the operad $d{\mathcal {O}}$, which are just $d{\mathcal {O}}$-algebras in $\operatorname {GrCh}^{\mathbb {S}}$, see [Reference Fresse17, §3.2.9]. They carry a semi-model category structure transferred along

\[ \operatorname{GrCh}^{\mathbb{S}}\mathop{\rightleftarrows}^{d{\mathcal{O}}\circ{-}}_{\text{forget}}\operatorname{Alg}_{\operatorname{GrCh}^{\mathbb{S}}}({d{\mathcal{O}}}), \]

see [Reference Fresse17, theorem 12.3.A]. The model structure on $\operatorname {GrCh}^{\mathbb {S}}$ is combinatorial with sets of generating (trivial) cofibrations inherited from the usual ones in $\operatorname {Ch}$.

Let us check that the Koszul complex is cofibrant as a left $d{\mathcal {O}}$-module. This will conclude the proof since any cofibrant left $d{\mathcal {O}}$-module is cofibrant in $\operatorname {GrCh}^{\mathbb {S}}$ by [Reference Fresse17, proposition 12.3.2] because $d{\mathcal {O}}$ is also cofibrant therein, see lemma 7.10 below. If we filter the Koszul complex $d{\mathcal {O}}\circ _{\zeta }{d{\mathcal {O}}}^{\unicode{x00A1}}$ by the weight of ${d{\mathcal {O}}}^{\unicode{x00A1}}$, the differential strictly decreases the filtration level. Hence $F_n(d{\mathcal {O}}\circ _\zeta {d{\mathcal {O}}}^{\unicode{x00A1}})$ is obtained from $F_{n-1}(d{\mathcal {O}}\circ _{\zeta }{d{\mathcal {O}}}^{\unicode{x00A1}})$ by freely attaching the $\mathbb {S}$-module $({d{\mathcal {O}}}^{\unicode{x00A1}})^{(n)}$, compare the proof of [Reference Fresse18, proposition 2.12]. The object $({d{\mathcal {O}}}^{\unicode{x00A1}})^{(n)}$ is cofibrant in $\operatorname {GrCh}^{\mathbb {S}}$ by lemma 7.4. This suffices to prove that the left $d{\mathcal {O}}$-module $d{\mathcal {O}}\circ _{\zeta }{d{\mathcal {O}}}^{\unicode{x00A1}}$ is cofibrant.

Proof. It suffices to notice that the generators of the $d{\mathcal {O}}$-ideal $I_{<0}$ must be in the kernel of $f$, since they have negative horizontal degree.

Proof. It suffices to check that the underlying graded complex of $I_{<0}$ has trivial homology. First, we filter $\Omega _{d{\mathcal {O}}}\operatorname {B}_{d{\mathcal {O}}}(A)$. Its underlying bigraded module is $\mathcal {O}\circ \mathcal {D}\circ {\mathcal {D}}^{\unicode{x00A1}}\circ {\mathcal {O}}^{\unicode{x00A1}}\circ A$, and we filter it according to the weight of ${\mathcal {O}}^{\unicode{x00A1}}$ and the horizontal degree of $A$. This filtration is bounded below and exhaustive. We have to check that this is a filtration in $\operatorname {GrCh}$, i.e. compatibility with the differential.

The differential of $\Omega _{d{\mathcal {O}}}\operatorname {B}_{d{\mathcal {O}}}(A)$ consists of three pieces. This first one is induced by the differential $d_0$ of the underlying graded complex of $A$, so it preserves the filtration. The second one, given by the perturbation $\partial ^{A}$ of $\operatorname {B}_{d{\mathcal {O}}}(A)$, is compatible with the filtration because, in (7.5), $d_k$ reduces the horizontal degree since $k>0$, and the weight of $\mu ^{(1)}$ is smaller than the weight of $\mu$ because $\mu ^{(2)}$ has positive weight. Actually, this piece strictly decreases the filtration level. The third part is induced by the differential of the Koszul complex of $d{\mathcal {O}}$. In this piece, the filtration has already been considered in the proof of theorem 4.10. Using also the computation therein we see that

\[ E^{0}(\Omega_{d{\mathcal{O}}}\operatorname{B}_{d{\mathcal{O}}}(A))=\mathcal{O}\circ(\mathcal{D}\circ_{\kappa}{\mathcal{D}}^{\unicode{x00A1}})\circ{\mathcal{O}}^{\unicode{x00A1}}\circ A, \]

with the differential coming from the Koszul complex of $\mathcal {D}$ and from $d_0$ of $A$.