# coequalizers and cofiber in a quasicategory

$\require{AMScd}$ Let $C$ be a stable $\infty$-category and let $f,f':X \to Y$ be two morphisms. Why is the coequalizer (in C, not hC) of $f,f'$ equal to the cofiber of $f-f'$? (This is claimed in the proof of Prop. 1.1.3.4 in Lurie's Higher Algebra). I could understand this claim if we were talking about coequalizers in an additive category $hC$, but my question is why does it hold in $C$?

We have shown $hC$ is an additive category. So if $[f']$ denotes the class of $f'$ in $hC$, we have can take its additive inverse $-[f']$, and choose a representative $-f'$ (a morphism in $C$) for this class. Then a representative for $f-f'$ is the composite of the top line of the diagram \begin{CD} X @>>> X \times X @>(f,-f')>> Y \times Y @>>> Y \coprod Y@>>>Y \\ @VVV &&&&&& @VVV\\ 0 @>>> @>>> @>>> @>>> W \end{CD} and let $W$ be the pushout (in $C$). Why is $W$ the coequalizer of $f, f'$? In other words, why is the diagram $X \rightrightarrows Y \to W$ a colimit diagram?

There is a standard way of rewriting coequalizers as pushouts: the coequalizer $C$ of $f, f' : X \to Y$ is the pushout

$$\begin{CD} X \amalg X @>(f,f')>> Y \\ @VVV @VVV \\ X @>>> C \end{CD}$$

(where the left vertical map is the fold map, given by the identity on each summand).

Then to see that $C$ is also the cofiber of $f-f'$, look at this diagram:

$$\begin{CD} X @>{\begin{pmatrix}1 \\-1\end{pmatrix}}>> X \amalg X @>{\begin{pmatrix}f & f'\end{pmatrix}}>> Y \\ @VVV @V{\begin{pmatrix}1 & 1\end{pmatrix}}VV @VVV \\ 0 @>>> X @>>> C \end{CD}$$

Here I've switched to more comfortable matrix notation for morphisms where you implicitly insert the equivalence $\alpha : X \amalg X \cong X \times X$ or its inverse in the morphisms where required. For example, ${\scriptstyle\begin{pmatrix}1 \\ -1\end{pmatrix}}$ means the composite $X \xrightarrow{(\mathrm{id}, -\mathrm{id})} X \times X \xrightarrow{\alpha^{-1}} X \amalg X$.

Notice that the composite across the top is $f - f'$, so if we show that the outer rectangle is a pushout, then $C$ will be the cofiber of $f-f'$.

Since the square on the right is a pushout, to show the rectangle is a pushout it is enough to show the square on the left is a pushout. This is done by using the shear map to turn it into a more obvious pushout square, namely, look at:

$$\begin{CD} X @>{\begin{pmatrix}1 \\ 0\end{pmatrix}}>> X \amalg X @>\begin{pmatrix}1 & 0 \\ -1 & 1\end{pmatrix}>> X \amalg X \\ @VVV @V{\begin{pmatrix}0 & 1\end{pmatrix}}VV @V{\begin{pmatrix}1 & 1\end{pmatrix}}VV \\ 0 @>>> X @>{1}>> X \end{CD}$$

You can check that the outer rectangle is precisely the square we are trying to show is a pushout, so we'll be done if we show each of these two squares is a pushout.

The one on the right is because both horizontal maps are equivalences: the inverse of the top one is the shear map ${\begin{pmatrix}1 & 0 \\ 1 & 1\end{pmatrix}}$ which is an equivalence in any additive category.

The square on the left is the (pointwise) coproduct of two easy pushout squares:

$$\begin{CD} X @>>> X \\ @VVV @VVV \\ 0 @>>> 0 \end{CD} \quad\quad \begin{CD} 0 @>>> X \\ @VVV @VVV \\ 0 @>>> X \end{CD}$$

• Thank you for such a detailed explanation! It seems quite ingenious to me as well. – usr0192 Oct 30 '15 at 11:11
• This is crazy, @user90219: I had completely forgotten I had already answered this question on MathOverflow two years ago! That time I got confused about how to prove the first left hand square was a pushout and Akhil Matthew gave the argument in a comment. I'm glad to see I learned the proof even if I forgot where I learned it. Also, apparently that time I used the row-vector convention for matrices instead... – Omar Antolín-Camarena Oct 30 '15 at 23:56
• Thanks for the link, Omar. I feel less stupid for being puzzled by the claim knowing that Mike Shulman also wondered why it was true as well. – usr0192 Oct 31 '15 at 10:05