@@ -250,9 +250,9 @@ From the previous proposition, we deduce the following very useful corollary.

\begin{tikzcd}

\displaystyle\coprod_{x \in E}F_x \ar[r]\ar[d]& E \ar[ddr,bend left]\ar[d]&\\

A \ar[drr,bend right,"\beta"']\ar[r]& G \ar[dr, dotted]&\\

&&B

&&B,

\ar[from=1-1,to=2-2,phantom,very near end,"\ulcorner"]

\end{tikzcd},

\end{tikzcd}

\]

where the arrow $E \to B$ is the canonical inclusion. Hence, by universal property, the dotted arrow exists and makes the whole diagram commutes. A thorough verification easily shows that the morphism $G \to B$ is a monomorphism of $\Rgrph$.

...

...

@@ -261,11 +261,11 @@ From the previous proposition, we deduce the following very useful corollary.

\begin{tikzcd}[row sep = large]

\displaystyle\coprod_{x \in E}F_x \ar[r]\ar[d]& E \ar[d]&\\

A \ar[d,"\alpha"]\ar[r]& G \ar[d]\ar[r]& B \ar[d,"\delta"]\\

C \ar[r]& H \ar[r]& D

C \ar[r]& H \ar[r]& D.

\ar[from=1-1,to=2-2,phantom,"\ulcorner" very near end,"\text{\textcircled{\tiny\textbf{1}}}" near start, description]

\ar[from=2-1,to=3-2,phantom,"\ulcorner" very near end,"\text{\textcircled{\tiny\textbf{2}}}", description]

\ar[from=2-2,to=3-3,phantom,"\ulcorner" very near end,"\text{\textcircled{\tiny\textbf{3}}}", description]

\end{tikzcd}.

\end{tikzcd}

\]

What we want to prove is that the image by the functor $L$ of the pasting of squares \textcircled{\tiny\textbf{2}} and \textcircled{\tiny\textbf{3}} is homotopy cocartesian. Since the morphism $G \to B$ is a monomorphism, we deduce from Corollary \ref{cor:hmtpysquaregraph} that the image by the functor $L$ of square \textcircled{\tiny\textbf{3}} is homotopy cocartesian. Hence, in virtue of Lemma \ref{lemma:pastinghmtpycocartesian}, all we have to show is that the image by $L$ of square \textcircled{\tiny\textbf{2}} is homotopy cocartesian. On the other hand, we know that both morphisms

\[

...

...

@@ -818,11 +818,16 @@ $2$\nbd{}categories, let us recall the following particular case of Corollary

Notice that $A_{(1,1)}$ is nothing but $\sD_2$. We are going to prove that if $n\neq0$ or $m\neq0$, then $A_{(m,n)}$ is \good{} and has the homotopy type of a point. When $m\neq0$\emph{and}$n\neq0$, this result is not surprising, but when $n=0$ or $m=0$ (but not both), it is \emph{a priori} less clear what the homotopy type of $A_{(m,n)}$ is and whether it is \good{} or not. For example, $A_{(1,0)}$ can be pictured as follows

Notice that $A_{(1,1)}$ is nothing but $\sD_2$. We are going to prove that if $n\neq0$ or $m\neq0$, then $A_{(m,n)}$ is \good{} and has the homotopy type of a point. When $m\neq0$\emph{and}$n\neq0$, this result is not surprising, but when $n=0$ or $m=0$ (but not both), it is \emph{a priori} less clear what the homotopy type of $A_{(m,n)}$ is and whether it is \good{} or not. For example, $A_{(1,0)}$ can be pictured as

\[

%% \begin{tikzcd}

%% A \ar[r, bend left=70, "f",""{name=A,below}] \ar[r,bend right=70,"1_A"',""{name=B,above}] & A, \ar[from=A,to=B,Rightarrow,"\alpha"]

%% \end{tikzcd}

%% \text{ or }

\begin{tikzcd}

A \ar[r, bend left=70, "f",""{name=A,below}]\ar[r,bend right=70,"1_A"',""{name=B,above}]& A \ar[from=A,to=B,Rightarrow,"\alpha"]

\end{tikzcd},

A \ar[loop,in=50,out=130,distance=1.5cm,"f",""{name=A,below}]

\ar[from=A,to=1-1,Rightarrow,"\alpha"]

\end{tikzcd}

\]

and has many non trivial $2$-cells, such as $f\comp_0\alpha\comp_0 f$.

...

...

@@ -856,9 +861,9 @@ For any $n \geq 0$, consider the following cocartesian square

\[

\begin{tikzcd}

\Delta_1\ar[r,"i"]\ar[d,"\tau"]&\Delta_n \ar[d]\\

A_{(1,1)}\ar[r]& A_{(1,n)}

A_{(1,1)}\ar[r]& A_{(1,n)},

\ar[from=1-1,to=2-2,phantom,very near end,"\ulcorner"]

\end{tikzcd},

\end{tikzcd}

\]

where $\tau : \Delta_1\to A_{(1,1)}$ is the $2$-functor that sends the unique non-trivial $1$\nbd{}cell of $\Delta_1$ to the target of the generating $2$-cell of $A_{(1,1)}$. It is not hard to check that $\tau$ is strong deformation retract and thus, a co-universal Thomason equivalence (Lemma \ref{lemma:pushoutstrngdefrtract}). Hence, the morphism $A_{(1,1)}\to A_{(1,n)}$ is also a (co-universal) Thomason equivalence and the square is Thomason homotopy cocartesian (Lemma \ref{lemma:hmtpycocartsquarewe}). Now, the morphism $\tau : \Delta_1\to A_{(1,1)}$ is also a folk cofibration and since $\Delta_1$, $\Delta_n$ and $A_{(1,1)}$ are \good{}, it follows from what we said in \ref{paragr:criterion2cat} that $A_{(1,n)}$ is \good{}. Finally, since $\Delta_1$, $\Delta_n$ and $A_{(1,1)}$ have the homotopy type of a point, the fact that the previous square is Thomason homotopy cocartesian implies that $A_{(1,n)}$ has the homotopy type of a point.

...

...

@@ -866,9 +871,9 @@ For any $n \geq 0$, consider the following cocartesian square

\ar[from=1-1,to=2-2,phantom,very near end,"\ulcorner"]

\end{tikzcd},

\end{tikzcd}

\]

where $\sigma : \Delta_1\to A_{(1,1)}$ is the $2$-functor that sends the unique non trivial $1$\nbd{}cell of $\Delta_1$ the source of the generating $2$\nbd{}cell of $A_{(1,1)}$, we can prove that $A_{(m,1)}$ is \good{} and has the homotopy type of a point.

...

...

@@ -1171,14 +1176,17 @@ isomorphisms, which means by definition that $P$, $P'$ and $P''$ are \good{}.

\begin{itemize}[label=-]

\item generating $0$\nbd{}cell: $A$,

\item generating $1$\nbd{}cell: $f,g: A \to A$,

\item generating $2$\nbd{}cell: $\alpha,\beta : f \to g$.

\item generating $2$\nbd{}cell: $\alpha,\beta : f \Rightarrow g$.

\end{itemize}

In picture, this gives:

%% \[

%% \begin{tikzcd}

%% \end{tikzcd}

%% \]

\[

\begin{tikzcd}[column sep=huge]

A \ar[r,bend left=75,"f",""{name=A,below,pos=9/20},""{name=C,below,pos=11/20}]\ar[r,bend right=75,"g"',""{name=B,above,pos=9/20},""{name=D,above,pos=11/20}]& A.

\ar[from=C,to=D,bend left,"\alpha",Rightarrow]

\ar[from=A,to=B,bend right,"\beta"',Rightarrow]

\end{tikzcd}

\]

\todo{À finir}

\end{paragr}

\begin{paragr}

Let $P$ be the free $2$\nbd{}category defined as follows:

...

...

@@ -1197,9 +1205,9 @@ isomorphisms, which means by definition that $P$, $P'$ and $P''$ are \good{}.

\ar[from=E,to=F,Rightarrow,"\gamma",bend left]

\end{tikzcd}

\]

Now let $P'$ be the sub-$2$\nbd{}category of $P$ spanned by $A$,$B$,

$\alpha$ and $beta$, and let $P''$ be the sub-$2$\nbd{}category of $P$

spanned by $A$,$B$,$\beta$ and $\gamma$. These $2$\nbd{}categories are

Now let $P'$ be the sub-$2$\nbd{}category of $P$ spanned by $A$,$B$,

$\alpha$ and $\beta$, and let $P''$ be the sub-$2$\nbd{}category of $P$

spanned by $A$,$B$,$\beta$ and $\gamma$. These $2$\nbd{}categories are

simply copies of $\sS_2$. Notice that we have a cocartesian

square

\begin{equation}\label{square:bouquet}

...

...

@@ -1242,30 +1250,41 @@ isomorphisms, which means by definition that $P$, $P'$ and $P''$ are \good{}.

\ar[from=G,to=H,Rightarrow,"\delta",bend left]

\end{tikzcd}

\]

Let us prove that this $2$\nbd{}category is \good{}. Let $P'$ be the sub-$2$\nbd{}category of $P$ spanned by $A$, $B$, $f$, $g$, $\alpha$ and $\beta$, let $P''$ be the sub-$2$\nbd{}category of $P$ spanned by $A$, $B$, $g$, $h$, $\gamma$ and $\delta$ and let $P''$ be the sub-$2$\nbd{}category of $P$ spanned by $A$, $B$ and $g$. These two$2$\nbd{}categories are copies of $\sS_2$ and we have a cocartesian square

Let us prove that this $2$\nbd{}category is \good{}. Let $P_0$ be the sub-$1$category of $P$ spanned by $A$, $B$ and $g$, let $P_1$ be the sub-$2$\nbd{}category of $P$ spanned by $A$, $B$, $g$, $h$, $\gamma$ and $\delta$ and let $P_2$ be the sub-$2$\nbd{}category of $P$ spanned by $A$, $B$, $f$, $g$, $\alpha$ and $\beta$. The$2$\nbd{}categories $P_1$ and $P_2$are copies of $\sS_2$, and $P_0$ is a copy of $\sS_1$. Moreover, we have a cocartesian square of inclusions

\[

\begin{tikzcd}

\sD_1\ar[d,"\langle g \rangle"]\ar[r,"\langle g \rangle"]& P'\ar[d]\\\

P''\ar[r]& P.

P_0\ar[d,hook]\ar[r,hook]& P_2\ar[d,hook]\\\

P_1\ar[r,hook]& P.

\end{tikzcd}

\]

Let us prove that this square is Thomason homotopy cocartesian using the second part of Corollary \ref{prop:critverthorThomhmtpysquare}. First, all the morphisms of the previous square are isomorphisms on objects and thus, the image by $V_0$ of the above square is obviously cocartesian. Now, notice that the categories $P(A,B)$, $P'(A,B)$ and $P''(A,B)$ are respectively free on the graphs

Let us prove that this square is Thomason homotopy cocartesian using the second part of Corollary \ref{prop:critverthorThomhmtpysquare}. This means that we have to show that for every $k \geq0$, the induced square

\[

\begin{tikzcd}

f \ar[r,shift left,"\alpha"]\ar[r, shift right, "\beta"']& g \ar[r,shift left,"\gamma"']\ar[r,shift right,"\delta"]& h,

\end{tikzcd}

\]

\[

\begin{tikzcd}

f \ar[r,shift left,"\alpha"]\ar[r, shift right, "\beta"']& g,

\end{tikzcd}

\]

and

\[

\begin{tikzcd}

g \ar[r,shift left,"\gamma"]\ar[r,shift right,"\delta"']& h.

H_k(P_0)\ar[d]\ar[r]& H_k(P_2)\ar[d]\\\

H_k(P_1)\ar[r]& H_k(P)

\end{tikzcd}

\]

is cocartesian.

First, all the morphisms of the previous square are isomorphisms on objects and thus, the image by $V_0$ of the above square is obviously cocartesian. Now, notice that all the categories $P_i(A,A)$ and $P_i(B,B)$ for $0\leq i \leq2$ are all isormophic to the terminal category $\sD_0$ and the categories $P_i(B,A)$ for $0\leq i \leq2$ are all the empty category. It follows that for

%% notice that the categories $P(A,B)$, $P'(A,B)$ and $P''(A,B)$ are respectively free on the graphs

%% \[

%% \begin{tikzcd}

%% f \ar[r,shift left,"\alpha"] \ar[r, shift right, "\beta"'] & g \ar[r,shift left,"\gamma"'] \ar[r,shift right,"\delta"] & h,

%% \end{tikzcd}

%% \]

%% \[

%% \begin{tikzcd}

%% f \ar[r,shift left,"\alpha"] \ar[r, shift right, "\beta"'] & g,

%% \end{tikzcd}

%% \]

%% and

%% \[

%% \begin{tikzcd}

%% g \ar[r,shift left,"\gamma"] \ar[r,shift right,"\delta"'] & h.

%% \end{tikzcd}

%% \]

Besides, all the categories $P(A,A)$, $P(B,B)$, $P'(A,A)$, $P'(B,B)$, $P''(A,A)$ and $P''(B,B)$ are all isormophic to the terminal category $\sD_0$.

@@ -52,7 +52,7 @@ The functor $\kappa : \Psh{\Delta} \to \Ch$ is left Quillen and sends weak equiv

as a generating trivial cofibrations (see for example \cite[Section I.1]{goerss2009simplicial} for the notations). A quick computation, which we leave to the reader, shows that the image by $\kappa$ of $\partial\Delta_n \hookrightarrow\Delta_n$ is a monomorphism with projective cokernel and the image by $\kappa$ of $\Lambda^i_n \hookrightarrow\Delta_n$ is a quasi-isomorphism. This proves that $\kappa$ is left Quillen. Since all simplicial sets are cofibrant, it follows from Ken Brown's Lemma \cite[Lemma 1.1.12]{hovey2007model} that $\kappa$ also preserves weak equivalences.

\end{proof}

\begin{remark}

The previous lemma admits also as more conceptual proof as follows. From the Dold-Kan equivalence, we know that $\Ch$ is equivalent to the category $\Ab(\Delta)$ of simplicial abelian groups and with this identification the functor $\kappa : \Psh{\Delta}\to\Ch$ is left adjoint of the canonical forgetful functor

The previous lemma admits also a more conceptual proof as follows. From the Dold-Kan equivalence, we know that $\Ch$ is equivalent to the category $\Ab(\Delta)$ of simplicial abelian groups and with this identification the functor $\kappa : \Psh{\Delta}\to\Ch$ is left adjoint of the canonical forgetful functor

\[

U : \Ch\simeq\Ab(\Delta)\to\Psh{\Delta}

\]

...

...

@@ -641,7 +641,7 @@ The following proposition is an immediate consequence of Theorem \ref{thm:cisins

\[

H^{\pol}_k(C)=\begin{cases}\mathbb{Z}\text{ if } k=0,2\\0\text{ in other cases.}\end{cases}

\]

On the other hand, it was proven in \cite[Theorem 4.9 and Example 4.10]{ara2019quillen} that (the nerve of) $C$ is a $K(\mathbb{Z},2)$-space. In particular, it has non-trivial singular homology groups in every even dimension. This proves that $\sH^{\pol}(C)$ is \emph{not} isomorphic to $\sH^{\sing}(C)$; which means that triangle \eqref{cmprisontrngle} cannot be commutative (up to an isomorphism).

On the other hand, it is proven in \cite[Theorem 4.9 and Example 4.10]{ara2019quillen} that (the nerve of) $C$ is a $K(\mathbb{Z},2)$-space. In particular, it has non-trivial singular homology groups in every even dimension. This proves that $\sH^{\pol}(C)$ is \emph{not} isomorphic to $\sH^{\sing}(C)$; which means that triangle \eqref{cmprisontrngle} cannot be commutative (up to an isomorphism).

\end{paragr}

Another consequence of the above counter-example is the following result, which we claimed in \ref{paragr:polhmlgythomeq}. Recall that given a morphism $u : C \to D$ of $\oo\Cat$, we write $\sH^{\pol}(u)$ instead of $\sH^{\pol}(\gamma^{\folk}(u))$.

The uniqueness of such a factorization follows from a similar argument which is left to the reader. This proves that $\overline{\sH^{\pol}}$ is the left derived functor of $\lambda$ when $\oo\Cat$ is equipped with Thomason equivalences and in particular we have

...

...

@@ -976,8 +976,8 @@ This is \cite[Theorem 5]{lafont2010folk}. (Although the part concerning generati

\[

\begin{tikzcd}

C \ar[d,"\eta_C"]\ar[r,"f"]& D \ar[d,"\eta_D"]\\

T(C)\ar[r,"T(f)"]& T(D)

\end{tikzcd},

T(C)\ar[r,"T(f)"]& T(D),

\end{tikzcd}

\]

where $\eta$ is the unit of the adjunction $\tau^{i}_{\leq n}\dashv\iota_n$.

...

...

@@ -1088,9 +1088,9 @@ We now turn to truncations of chain complexes.

\[

\begin{tikzcd}

\tau^{i}_{\leq n}(A)\ar[r]\ar[d,"\tau^{i}_{\leq n}(j)"']& X \ar[d,"g"]\\

\tau^{i}_{\leq n}(B)\ar[r]& Y

\tau^{i}_{\leq n}(B)\ar[r]& Y,

\ar[from=1-1, to=2-2, phantom, "\ulcorner",very near end]

\end{tikzcd},

\end{tikzcd}

\]

the morphism $\iota_n(g)$ is a weak equivalence of $\Ch$. As explained in \cite[Proposition 7.19]{dwyer1995homotopy}, there exists a set of generating trivial cofibrations of the projective model structure on $\Ch$ consisting of the maps

From Proposition \ref{prop:singhmlgylowdimension}, we know that $\alpha^{\sing}$ induces isomorphisms \[H_k^{\sing}(C)\simeq H_k(\lambda(C))\] for $k \in\{0,1\}$ and from Corollary \ref{cor:polhmlgycofibrant} and Paragraph \ref{paragr:polhmlgylowdimension} we know that $\alpha^{\pol}$ induces isomorphisms $H_k^{\pol}(C)\simeq H_k(\lambda(C))$ for $k \in\{0,1\}$. The result follows then from an immediate 2-out-of-3 property.

\section{Tensor product and oplax transformations}

...

...

@@ -271,9 +271,9 @@ From now on, we will consider that the category $\Psh{\Delta}$ is equipped with

\[

\begin{tikzcd}

X\ar[rd,"u"]\ar[d,"i_0^X"']&\\

\sD_1\otimes X \ar[r,"\alpha"]& Y \\

X \ar[ru,"v"']\ar[u,"i_1^X"]

\end{tikzcd},

\sD_1\otimes X \ar[r,"\alpha"]& Y,\\

X \ar[ru,"v"']\ar[u,"i_1^X"]&

\end{tikzcd}

\]

where $i_0^X$ and $i_1^X$ are induced by the two $\oo$-functors $\sD_0\to\sD_1$ and where we implicitly used the isomorphism $\sD_0\otimes X \simeq X$, is commutative.

\item As an $\oo$-functor $\alpha : X \to\homlax(\sD_1,Y)$ such that the following diagram

...

...

@@ -410,15 +410,15 @@ From now on, we will consider that the category $\Psh{\Delta}$ is equipped with

\begin{equation}\label{diagramtransf}\tag{ii}

\begin{tikzcd}

B\ar[rd,"\mathrm{id}_B"]\ar[d,"i_0^B"']&\\

\sD_1\otimes B \ar[r,"\alpha"]& B \\

B \ar[ru,"i\circ r"']\ar[u,"i_1^B"]

\end{tikzcd},

\sD_1\otimes B \ar[r,"\alpha"]& B,\\

B \ar[ru,"i\circ r"']\ar[u,"i_1^B"]&

\end{tikzcd}

\end{equation}

and

\begin{equation}\label{diagramstrong}\tag{iii}

\begin{tikzcd}

\sD_1 \otimes A \ar[rr, bend right,"p\otimes i"']\ar[r,"\sD_1 \otimes i"]&\sD_1 \otimes B \ar[r,"\alpha"]& B

\end{tikzcd},

\sD_1 \otimes A \ar[rr, bend right,"p\otimes i"']\ar[r,"\sD_1 \otimes i"]&\sD_1 \otimes B \ar[r,"\alpha"]& B,

\end{tikzcd}

\end{equation}

where $p$ is the unique morphism $\sD_1\to\sD_0$, are commutative.

...

...

@@ -427,9 +427,9 @@ From now on, we will consider that the category $\Psh{\Delta}$ is equipped with

\begin{tikzcd}

A \ar[r,"u"]\ar[d,"i"]& A' \ar[d,"i'"]\ar[dd,bend left=75,"\mathrm{id}_{B'}"]\\

B \ar[d,"r"]\ar[r,"v"]& B' \ar[d,"r'",dashed ]\\

A \ar[r,"u"]& A'

A \ar[r,"u"]& A',

\ar[from=1-1,to=2-2,phantom,"\ulcorner",very near end]

\end{tikzcd},

\end{tikzcd}

\]

we deduce the existence of $r' : B' \to A'$ that makes the whole diagram commutes. In particular, we have $r' \circ i' =\mathrm{id}_{B'}$.

...

...

@@ -438,8 +438,8 @@ From now on, we will consider that the category $\Psh{\Delta}$ is equipped with

\begin{tikzcd}

\sD_1\otimes A \ar[r,"\sD_1\otimes u"]\ar[d,"\sD_1\otimes i"]&\sD_1\otimes A' \ar[d,"\sD_1\otimes i'"]\ar[dd,bend left=75,"p\otimes i'"]\\

\sD_1\otimes B \ar[d,"\alpha"]\ar[r,"\sD_1\otimes v"]&\sD_1\otimes B' \ar[d,"\alpha'",dashed ]\\

\sD_1\otimes B \ar[r,"v"]&\sD_1\otimes B'

\end{tikzcd}.

\sD_1\otimes B \ar[r,"v"]&\sD_1\otimes B'.

\end{tikzcd}

\]

The existence of $\alpha' : \sD_1\otimes B' \to B'$ that makes the whole diagram commutes follows from the fact that the functor $\sD_1\otimes\shortminus$ preserves colimits. In particular, we have $\alpha' \circ(\sD_1\otimes i')= p \otimes i'$.

@@ -200,8 +200,8 @@ We now turn to the most important way of obtaining op-prederivators.

and every $\begin{tikzcd}A \ar[r,bend left,"u",""{name=A,below}]\ar[r,bend right, "v"',""{name=B,above}]& B \ar[from=A,to=B,Rightarrow,"\alpha"]\end{tikzcd}$ induces by pre-composition a $2$-morphism of localizers

(This last property is trivial since a $2$-morphism of localizers is simply a natural transformation between the underlying functors.) Then, by the universal property of localization, we have for every $u : A \to B$ an induced functor, which we still denote $u^*$,

\[

...

...

@@ -210,8 +210,8 @@ We now turn to the most important way of obtaining op-prederivators.

and for every $\begin{tikzcd}A \ar[r,bend left,"u",""{name=A,below}]\ar[r,bend right, "v"',""{name=B,above}]& B \ar[from=A,to=B,Rightarrow,"\alpha"]\end{tikzcd}$, an induced natural transformation

For example, when $\sD$ is the homotopy op-prederivator of a localizer, $B$ is the terminal category $e$, for any $X$ object of $\sD(A)$ the previous canonical morphism reads

\[

...

...

@@ -527,8 +527,8 @@ We now turn to the most important way of obtaining op-prederivators.

\[

\begin{tikzcd}

(0,0)\ar[d]\ar[r]&(0,1)\\

(1,0)&

\end{tikzcd}.

(1,0)&.

\end{tikzcd}

\]

Finally, we write $i_{\ulcorner} : \ulcorner\to\square$ for the canonical inclusion functor.

\end{paragr}

...

...

@@ -559,17 +559,17 @@ We now turn to the most important way of obtaining op-prederivators.

Suppose now that $\sD$ has left Kan extensions. For $X$ an object of $\sD(\square)$, we have a canonical morphism $p_!(i_{\ulcorner}^*(X))\to X_{(1,1)}$ defined as the composition

\[

...

...

@@ -582,8 +582,8 @@ We now turn to the most important way of obtaining op-prederivators.

X=

\begin{tikzcd}

A \ar[r,"u"]\ar[d,"f"]& B \ar[d,"g"]\\

C \ar[r,"v"]&D

\end{tikzcd},

C \ar[r,"v"]&D,

\end{tikzcd}

\]

this previous morphism reads

\[

...

...

@@ -622,8 +622,8 @@ We now turn to the most important way of obtaining op-prederivators.

\[

\begin{tikzcd}

A \ar[r,"u"]\ar[d,"1_A"]& B \ar[d,"1_B"]\\

A \ar[r,"u"]&B

\end{tikzcd}.

A \ar[r,"u"]&B.

\end{tikzcd}

\]

The result follows then from \cite[Proposition 3.12(2)]{groth2013derivators}.

For an $\oo$-category $C$ and $n\geq0$, an $\oo$-functor

...

...

@@ -405,11 +405,11 @@ So far, we have not yet seen examples of free $\oo$-categories. In order to do t

\[

(a \ast b)\ast(c \ast d)=(a \ast c )\ast(b \ast d).

\]

It is straightforward to see that a necessary and sufficient condition for this equation to hold is to require that $M$ be commutative. Hence, we have proven the following lemma.

It is straightforward to see that a necessary and sufficient condition for this equation to hold is to require that $M$ be commutative. Hence, we have proved the following lemma.