2
$\begingroup$

$\require{AMScd}$ 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?

$\endgroup$
2
$\begingroup$

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:

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

$\endgroup$
3
  • $\begingroup$ Thank you for such a detailed explanation! It seems quite ingenious to me as well. $\endgroup$ – usr0192 Oct 30 '15 at 11:11
  • 1
    $\begingroup$ 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... $\endgroup$ – Omar Antolín-Camarena Oct 30 '15 at 23:56
  • $\begingroup$ 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. $\endgroup$ – usr0192 Oct 31 '15 at 10:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.