# Homotopy pullbacks and pushouts in stable model categories

There are lots of similar questions that have been answered on this topic (particularly Homotopy limit-colimit diagrams in stable model categories), but I have a specific question that I do not believe has been answered. I suspect there is a simple answer that I'm just not seeing, so apologize in advance if this is the case. Given a commutative square $$\require{AMScd}$$ $$\begin{CD} A @>{}>> X\\ @V{f}VV @V{g}VV \\ B @>{}>> Y \end{CD}$$ we say it is a homotopy pullback square if the canonical map $$A \to holim(B \to Y \leftarrow X)$$ is a weak equivalence and that it is a homotopy pushout square if the canonical map $$hocolim(B \leftarrow A \rightarrow X) \to Y$$ is a weak equivalence.

I'm hoping to understand the proof that in a stable category, the two notions coincide. As pointed out in the reference above, in the 2007 version of Hovey's Model Categories,'' an argument is given in Remark 7.1.12 that goes roughly as follows. One can show that the square above is a homotopy pullback square if and only if $$hofib(f) \overset{\sim}{\to} hofib(g)$$ is a weak equivalence. Similarly, it is a homotopy pushout square if and only if $$hocofib(f) \overset{\sim}{\to} hocofib(g)$$ is a weak equivalence. One then concludes the result by claiming that $$\Sigma hofib(f) \simeq hocofib(f) \text{ and } \Sigma hofib(g) \simeq hocofib(g)$$ This is the part on which I am stuck. Hovey claims that it follows from Thereom 7.1.11 (which is the same in both the 1999 and 2007 version) but I do not see this. It seems to me (and here is where I must be going wrong) that Hovey is claiming that the fiber in the homotopy category is the homotopy fiber -- because from what I understand, it is the homotopy category which is triangulated (to which 7.1.11 would apply), not the original stable model category. Since the fiber in the homotopy category is not (usually) the homotopy fiber, I must be misinterpreting what Hovey is saying.

So my question is: how do we know that the suspension of the homotopy fiber is weakly equivalent to the homotopy cofiber (or, equivalently, that loops of the homotopy cofiber is weakly equivalent to the homotopy fiber). Am I misunderstanding what Hovey means by "fiber sequence?" Are the distinguished triangles actually homotopy cofiber sequences in the original model category? As is probably clear by now, I do not know much about triangulated categories. Any help/reference would be appreciated!

• For Hovey, (co)fiber sequences are homotopy (co)fiber sequences. This is stated explicitly in Definition 6.2.6. – Dmitri Pavlov Jul 9 at 1:24
• By Definition 6.2.6, a (co)fiber sequence in Ho(C) by definition comes from a (co)fiber sequence in C. So (co)fiber sequences in the homotopy category by definition lift to the original category. – Dmitri Pavlov Jul 9 at 16:41
• Yes, your first claim is correct (provided that suspension is understood in the derived sense). For the question about g: in general, you must compute the homotopy fiber and cofiber using two different resolutions of g. To compute the homotopy fiber of g, replace g by a weakly equivalent fibration between fibrant objects, then take the ordinary fiber F. To compute the homotopy cofiber of g, replace g by a cofibration between cofibrant objects, then take the ordinary cofiber. – Dmitri Pavlov Jul 10 at 22:14
• Then to compute the derived suspension of F, replace F by a weakly equivalent cofibrant object and compute its ordinary suspension S. Then S is weakly equivalent to the homotopy fiber of g computed above. – Dmitri Pavlov Jul 10 at 22:15
• The above sequences are constructed in the category C and use different replacements of g in C, but once we pass to Ho(C), these replacements become isomorphic. In this context, Hovey only talks about (co)fiber sequences in Ho(C), not C. Theorem 7.1.11, in particular, is formulated in Ho(C), so there is no problem in using two different resolutions of g. (Hovey uses the notation S=Ho(C).) – Dmitri Pavlov Jul 10 at 22:51

## 1 Answer

Many thanks and much credit to Dmitri Pavlov for clearing up my confusions. In the interest of having a definitive answer, I've organized his comments and my understanding into the following answer.

Let $$\mathcal{C}$$ be a stable model category; then (by definition, for Hovey) $$Ho(\mathcal{C})$$ is a triangulated category.

Note: By definition 6.2.6 (same in both the 1999 and 2007 version of Hovey's "Model Categories") when we say $$A \to B \to C$$ in $$Ho(\mathcal{C})$$ is a cofiber sequence, we mean that there is a natural zig-zag of weak equivalences in $$\mathcal{C}$$ to a cofibration sequence of cofibrant objects $$A' \hookrightarrow B' \to C'$$. Dually, saying that $$X \to Y \to Z$$ in $$Ho(\mathcal{C})$$ is a fiber sequence means that there is a natural zig-zag of weak equivalences in $$\mathcal{C}$$ to a fibration sequence of fibrant objects $$X' \to Y' \twoheadrightarrow Z'$$.

Claim: Given a map $$X \overset{g}{\to} Y$$ in $$\mathcal{C}$$, we have that $$\Sigma hofib(g) \simeq hocofib(g)$$, where $$\Sigma$$ is derived suspension (i.e. the suspension of a cofibrant replacement).

Proof: Without loss of generality (replacing if needed) we may assume $$g$$ is a fibration between fibrant objects. Take the (strict) fiber of $$g$$ (which will also be the homotopy fiber) to obtain $$F \to X \overset{g}{\twoheadrightarrow} Y$$ Then replace the map $$F \to X$$ with a fibration and again take the fiber to obtain the following fibration sequence of fibrant objects in $$\mathcal{C}$$ $$\Omega Y \to G \twoheadrightarrow X \twoheadrightarrow Y$$ By definition, this descends to a fiber sequence $$\Omega Y \to G \to X \to Y$$ in $$Ho(\mathcal{C})$$. By Remark 7.1.9 (same in both versions) the following is also then a fiber sequence in $$Ho(\mathcal{C})$$ (note our notation clashes with Hovey's here, i.e. our $$G$$ is his $$X$$, etc.). $$\Omega\Sigma G \to X \to Y \to \Sigma G$$ which means (by 6.2.6) that $$hofib(X \to Y) \simeq \Omega\Sigma G$$. Now apply 7.1.11 (same in both versions) in $$Ho(\mathcal{C})$$ (taking $$Z = \Sigma G$$) to obtain a cofiber sequence $$\Omega \Sigma G \to X \to Y \to \Sigma(\Omega\Sigma G)$$ in $$Ho(\mathcal{C})$$ which means (by 6.2.6) that $$hocofib(X \overset{g}{\to} Y) \simeq \Sigma(\Omega\Sigma G)$$. Since $$\Sigma(\Omega\Sigma G) \simeq \Sigma hofib(X \overset{g}{\to} Y)$$, this completes the proof.