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Let $f,g\colon X\rightarrow Y$ be maps between spaces.

I am looking for criteria for the equalizer of $f$ and $g$ to be a homotopy equalizer and I am happy to get answers for whatever model category of spaces you prefer.

As the equalizer of $f$ and $g$ can be written as the pullback of $(f,g)\colon X\rightarrow Y\times Y$ along the diagonal $Y\rightarrow Y\times Y$, it is a homotopy equalizer if $(f,g)$ is a fibration, which unfortunately doesn't happen very often.

I feel there should be something weaker which is more likely to happen. An indication for this is the following thought. If $k\colon Z\rightarrow Y$ and $l\colon W\rightarrow Y$, the pullback of $k$ and $l$ is the equalizer of the maps from $Z\times W$ to $Y$ given by projecting to one factor first and then using $k$ respectively $l$. The map analogous to $(f,g)$ is then $k\times l$. But $k\times l$ is often not a fibration even if the pullback is a homotopy pullback, e.g. if only one of the maps is a fibration and the other is not.

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    $\begingroup$ Write the homotopy equalizer $hE(f,g)$ as the homotopy pullback of $X\xrightarrow{\Delta} X\times X\xleftarrow{(p,q)} N(f,g)$ Where $N(f,g)$ is the homotopy pullback of $f$ and $g$ and $(p,q):N(f,g)\rightarrow X\times X$ are the canonical maps. When $f$ or $g$ is a fibration then $N(f,g)$ is homotopy equivalent to the strict pull back $P(f,g)$ of the two maps. When both $f$ and $g$ are fibrations, then the map $(p,q):N(f,g)\rightarrow X\times X$ is a fibration so therefore $hE(f,g)$ is homotopy equivalent to the strict pullback of the above and this is precisely the equalizer $E(f,g)$. $\endgroup$
    – Tyrone
    Jun 24, 2016 at 16:40
  • $\begingroup$ This is an answer. $\endgroup$
    – Effy
    Jun 28, 2016 at 16:47
  • $\begingroup$ @Tyrone Why is the map $(p,q): N(f,g) \rightarrow X \times X$ a fibration? Take $X = Y$ and $f,g$ to be identity then the $(p,q)$ is just the diagonal map which is not a fibration. $\endgroup$
    – iron feliks
    May 24, 2018 at 11:21
  • $\begingroup$ $N(f,g)$ is the homotopy pullback, not strict pullback. $\endgroup$
    – Tyrone
    May 24, 2018 at 11:43
  • $\begingroup$ I still don't understand; Assume model category is right proper, then if either of $f$ or $g$ is a fibration then homotopy pullback is same (weakly equivalent) as strict pullback $\endgroup$
    – iron feliks
    May 24, 2018 at 11:52

1 Answer 1

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It is not sufficient that $f$ and $g$ are both fibrations.

Here is a counterexample in the category $\mathbf{Cat}$. The counterexample works both in the Thomason model structure (which gives a model category of spaces) as well as the canonical model structure.

Let $I$ be the "walking isomorphism": the category with two objects that is equivalent to $1$.

Let $F$ and $G$ be the two automorphisms of $I$. These are isomorphisms, and thus (acyclic) fibrations.

Since $I$ is contractible, every homotopy equalizer of $F$ and $G$ is also a contractible category.

The strict equalizer of $F$ and $G$ is the empty category.

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