# Coequalizers in stable (infinity,1)-categories

I have read it claimed in several places that in a stable $(\infty,1)$-category, the coequalizer of parallel maps $f,g:X\to Y$ can be identified with the cokernel of $f-g$ (i.e. the pushout of the map $X\to 0$ along $f-g$). How is this proven? Is it written down anywhere?

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For the last question, my guess is no, it's not written down anywhere (that's only a guess mind you). – Omar Antolín-Camarena Jul 23 '13 at 13:14

It seems to me like the proof for Abelian categories works here too. Consider the following diagram where $d = (id,-id)$, the middle vertical map is the fold (or $+$) map, and the square on the right is a homotopy pushout:

$$\begin{array}{ccccc} X & \xrightarrow{d} & X \oplus X & \xrightarrow{(f,g)} & Y \\ \downarrow & & \downarrow & & \downarrow \\ 0 & \to & X & \to & C \\ \end{array}$$

From the second square being a pushout, $C$ is the coequalizer of $f$ and $g$. If we check the first square is a pushout too, we get that the outer rectangle is a pushout and so $C$ is also $\mathrm{cofib}(f-g)$.

Here's Akhil's proof that the left hand square is a pushout:

$$\begin{array}{ccccc} X & \xrightarrow{\pmatrix{0&1}} & X \oplus X & \xrightarrow{\pmatrix{0 & 1 \\ 1 & -1}} & X \oplus X \\ \downarrow & & \downarrow \rlap{\scriptstyle\pmatrix{1\\ 0}}& & \downarrow \rlap{\scriptstyle\pmatrix{1\\1}}\\ 0 & \to & X & \xrightarrow{1} & X \\ \end{array}$$

The outer rectangle is what we'd like to show is a pushout. The right hand square is a pushout because both horizontal arrows are isomorphisms, and the left hand square is a pushout because it is the (pointwise) direct sum of these two (where all the maps $X \to X$ are the identity):

$$\begin{array}{cccccc} 0 & \to & X & & X & \to & X \\ \downarrow & & \downarrow & & \downarrow & & \downarrow \\ 0 & \to & X & & 0 & \to & 0 \\ \end{array}$$

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Hi Omar. I think you can use the "shearing" isomorphism of $X \oplus X$ to turn the sum map $X \oplus X \to X$ into the map $(1, 0): X \oplus X \to X$, and now the map $d$ turns into $(0, 1): X \to X \oplus X$, so the left-hand exact triangle becomes a sum of two (simpler) exact triangles. – Akhil Mathew Jul 22 '13 at 18:45
I don't think you need to do anything this complicated in an ordinary abelian category; can't you just say that since composition distributes over addition, a map $h:Y\to Z$ satisfies $h f = h g$ iff it satisfies $h (f - g) = 0$, so the two have the same universal property? – Mike Shulman Jul 23 '13 at 17:03
As for this proof, how do you show that $C$ being a pushout of that square is equivalent to being the coequalizer? Also, I think I know the answer to this, but how do you actually obtain all of these diagrams (which presumably are supposed to be maps $\Delta^1 \times \Delta^2\to \mathcal{C}$ or something)? – Mike Shulman Jul 23 '13 at 17:09
To show that $C$ is the coequalizer, I'd say that you just take mapping spaces into an arbitrary object $Z$ and that reduces it to showing that for spaces a homotopy equalizer of two maps $p,q:A\to B$ can be computed as the homotopy pullback of $B \xrightarrow{diag} B \times B \xleftarrow{(p,q)} A$. – Omar Antolín-Camarena Jul 23 '13 at 17:55
Of course, $\Delta^1\times \Delta^2$ is more than just four 2-simplices, but I guess four 2-simplices includes into it by an anodyne map or something? – Mike Shulman Jul 23 '13 at 19:14

I think you prove this pretty much as one does in an ordinary abelian category.

First note that the coequalizer $f,g: X \rightarrow Y$ is also given by the pushout of $(f, 1)$ and $(g, 1)$, both maps $X \oplus Y \rightarrow Y$. (This is proven exactly as in the ordinary case, by checking universal properties).

So it suffices to identify this pushout with the other. For this, note that the pushout of $X \rightarrow 0$ along the map $X \rightarrow X \oplus Y$ given by $(1, -g)$ is $Y$ together with the map $X \oplus Y \rightarrow Y$ given by $(g, 1)$ (since $g-g = 0$). Now use the pasting lemma for pushouts (which works in an $\infty$-categorical context, Lemma 4.4.2.1 in HTT).

Here's a few more details if you don't appreciate how reckless I'm being: For the first step it suffices by Lemma 4.2.4.3 of HTT to show that the map $\text{Fun}(\text{Fork}, \mathcal{C}) \rightarrow \text{Fun}(\Delta^1 \vee \Delta^1, \mathcal{C})$ given by the construction I've outlined above, is fully faithful. That way the object representing the functor assigning to an object the space of maps out of a fork diagram or a deleted square corresponding to it will be represented by equivalent objects, by Yoneda. To prove this, note that we have an inverse functor defined on the essential image which is an equivalence by the universal property of a coproduct, then the result follows by 2 out of 3.

To show that the pushout of $X \rightarrow 0$ by $X \rightarrow X \oplus Y$ is as described we will use another pasting argument. First note that the cofiber of the anti-diagonal $$X \stackrel{(1,-1)}\rightarrow X \oplus X$$ is just $X$ where the map $X \oplus X \rightarrow X$ is addition. Using this pushout square and the fact that the pushout of $+: X \oplus X \rightarrow X$ along $$X \oplus X \stackrel{(1, g)}\rightarrow Y$$ is $Y$ with the desired map, we can apply the pasting lemma and we're done.

I think I can make this last part a little more rigorous using the characterization of the additive inverse of a morphism given by Lemma 1.1.2.10 in Higher Algebra. If you like I can add that in later...

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For me, in the ordinary case, identifying the coequalizer with that pushout "by checking universal properties" would go like "the pushout represents pairs $h,k:Y\to Z$ such that $h(g,1) = k(f,1)$, i.e. $h g = k f$ and $h = k$, hence just a single map $h:Y\to Z$ such that $h g = h f$." I very much appreciate your word "reckless" for calling any $(\infty,1)$-categorical proof "the same" as this. (-: – Mike Shulman Jul 23 '13 at 17:13
Can you explain how your "construction" gives a map $\mathrm{Fun}(\mathrm{Fork},\mathcal{C}) \to \mathrm{Fun}(\Delta^1\vee\Delta^1,\mathcal{C})$ and not just an operation on objects? And what exactly is this "inverse functor"? Finally, yes, I would very much appreciate anything to make this more rigorous. – Mike Shulman Jul 23 '13 at 17:20