It is shown in Remark 7.1.12 of (a newer version of) Mark Hovey's book Model Categories that, in a stable model category, homotopy pullback squares coincide with homotopy pushout squares. The argument goes as follows: given a square $S$ of the form $$ \begin{matrix} W & \to & X\\ \downarrow_h & & \downarrow_g\\ Z & \to & Y \end{matrix} $$ one shows that $S$ is a pullback square if and only if the induced map $\mathrm{HoFibre}(h)\to\mathrm{HoFibre}(g)$ is an isomorphism in the homotopy category. Similarly, $S$ is a pushout square if and only if the induced map $\mathrm{HoCofibre}(h)\to\mathrm{HoCofibre}(g)$ is an isomorphism in the homotopy category. One concludes by observing that $\mathrm{HoCofibre}(h)$ is the suspension of $\mathrm{HoFibre}(h)$ and similarly for $g$.

Question: Is it possible to generalize this statement to say something like as follows?

"In a diagram of the form $$ \begin{matrix} W & & \to && X\\ &\searrow & \\ \downarrow &&Y&&\downarrow\\ &&&\searrow\\ Z & &\to && V, \end{matrix} $$ $V$ is a homotopy colimit of $$ \begin{matrix} W & & \to && X\\ &\searrow & \\ \downarrow &&Y&&\\ &&&\\\ Z & & && \end{matrix} $$ if and only if $W$ is a homotopy limit of $$ \begin{matrix} & & && X\\ & & \\ &&Y&&\downarrow\\ &&&\searrow\\ Z & &\to && V." \end{matrix} $$


2 Answers 2


This is not true. For example, take $W = X = Y = 0$, and the map $Z \rightarrow V$ to be an isomorphism. Then you've got a homotopy colimit diagram, but it is only a homotopy limit diagram if $Z$ is weakly contractible.

  • 2
    $\begingroup$ Why is it only a homotopy limit for weakly contractible $Z$ (or why doesn't the same counterexample work with two objects in the middle)? $\endgroup$
    – Rasmus
    Jul 1, 2013 at 23:18
  • $\begingroup$ Nevermind, Fernando explained it. Many thanks for your help! $\endgroup$
    – Rasmus
    Jul 2, 2013 at 8:43

Yes, because both statements are equivalent to the existence of an exact triangle of the form

$$W\longrightarrow X\oplus Y\oplus Z\longrightarrow V\longrightarrow\Sigma W.$$

The arrows are the same as in your diagram, except that you have to change of the sign of $W\rightarrow Y$ so that the first two arrows compose to $0$.

EDIT: The above 'answer' is shamely false. Let me just offer here a brief explanation of Jacob Lurie's correct answer. If the diagram is a homotopy colimit then we have an exact triangle

$$W\oplus W\stackrel{ \left(\begin{array}{cc} f&0\\ -g&g\\ 0&-h \end{array}\right) }\longrightarrow X\oplus Y\oplus Z\stackrel{(f',g',h')}\longrightarrow V\longrightarrow\Sigma(W\oplus W).$$

The homotopy limit condition gives rise to an exact triangle of the form

$$W\stackrel{ \left(\begin{array}{c} f\\ g\\ h \end{array}\right) }\longrightarrow X\oplus Y\oplus Z\stackrel{\left(\begin{array}{cc} f'&-g'&0\\ 0&g'&-h' \end{array}\right)}\longrightarrow V\oplus V\longrightarrow\Sigma W.$$

If $W=X=Y=0$ then you get an exact triangle by putting $V=Z$ and $h'=1_Z$ $$0\longrightarrow Z\stackrel{1}\longrightarrow Z\longrightarrow 0$$

But if you put $X=Y=0$, $V=Z$, and $h'=1_Z$, then $W=\Sigma^{-1}Z$ since the exact triangle is

$$\Sigma^{-1}Z\stackrel{0}\longrightarrow Z\stackrel{\binom{0}{1}}\longrightarrow Z\oplus Z\stackrel{(1,0)}\longrightarrow Z.$$

  • $\begingroup$ You are right, of course. Once more, thank you very much for your help! $\endgroup$
    – Rasmus
    Jul 1, 2013 at 21:46
  • $\begingroup$ You're very welcome! $\endgroup$ Jul 1, 2013 at 21:47
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    $\begingroup$ I think here is a problem: for two objects in the middle, we could make two equal maps sum up to zero by putting a minus sign on one. But we can never make three maps sum up to zero in this way. $\endgroup$
    – Rasmus
    Jul 1, 2013 at 23:05
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    $\begingroup$ Thank you so much for these explanations! Now I understand. Apparently I had a temporary misconception of how (co)limits actually look like and managed to convince you of it for a moment. $\endgroup$
    – Rasmus
    Jul 2, 2013 at 8:35
  • $\begingroup$ No, of course, it's the correct one! $\endgroup$ Jul 2, 2013 at 12:15

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