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As far as I know, one way to take a homotopy colimit in a model category is to replace (up to acyclic fibration) all arrows in the diagram with cofibrations, and take the strict colimit of the resulting diagram.

In Top with the model structure given by Serre fibrations, cofibrations, and weak equivalences, if one wants to obtain a homotopy pushout of the diagram $X \leftarrow A \rightarrow Y$, it is "enough" to replace only one of these arrows with a cofibration: that is, there is a natural map (by the universal property of the pushout) $Cyl(X)\cup_A Cyl(Y) \to Cyl(X) \cup_A Y$ that is a homotopy equivalence of spaces.

Question 1: What conditions on the model category $\mathcal{C}$ (or objects $X,Y,A$) will guarantee that the natural map $Cyl(X) \cup_A Cyl(Y) \to Cyl(X) \cup_A Y$ is a weak equivalence?

Question 2: This question is less precise, but if the map above is a weak equivalence, does that mean $Cyl(X) \cup_A Y$ is a good model for the homotopy pushout?

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    $\begingroup$ Your first sentence is not accurate, except when the indexing category of the colimit is of a rather special type, such as in the case of pushouts. For instance, homotopy orbits for a discrete group G are a homotopy colimit indexed on the category with one object with automorphisms G, and the identity map of any object X is a cofibration, but the homotopy orbit space X_hG for the trivial action is not X in general (in Spaces, it's X x BG). $\endgroup$
    – Reid Barton
    Commented Jan 8, 2010 at 20:33
  • $\begingroup$ Would the statement be accurate if the diagram category if the diagram D satisfied: the nerve of D is contractible? $\endgroup$
    – Joey Hirsh
    Commented Jan 8, 2010 at 20:41
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    $\begingroup$ No. It's not even possible to "replace all arrows in the diagram with cofibrations" for most indexing categories. For instance, consider Delta^op, the small category which indexes simplicial objects. If F: Delta^op --> Top, is a functor which takes every arrow to a cofibration, then in particular it takes every arrow to an injection. But it turns out that any such functor must be constant (Delta^op is generated by arrows which have either left or right inverses, so such an F has to take all these to isomorphisms.) $\endgroup$
    – Charles Rezk
    Commented Jan 8, 2010 at 21:22
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    $\begingroup$ For a discussion of why "replace all maps with cofibrations" is not always the way to obtain a cofibrant diagram, see Strom, Modern Classical Homotopy Theory, section 6.4.2, paragraph "A common misconception". He defines what it means for a small category to be "tree-like", and proves that in this case it is true that a diagram is cofibrant iff all the arrows are cofibrant. $\endgroup$
    – Bruno Stonek
    Commented Sep 29, 2016 at 11:35

2 Answers 2

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Question 1: The model category $\mathcal{C}$ should be left proper, i.e. the pushout of a weak equivalence along a cofibration is again a weak equivalence. (Dually, there is a notion of right proper.) Top is left proper, as is any model category in which every object is cofibrant, such as SSet. There is some information on this notion of properness at the nlab, and I think it's also discussed more thoroughly in Hirschhorn's book Model Categories and their Localizations (and probably many other places).

Question 2: Yes. People often say that a square in a model category is a homotopy pushout square if the induced map from the (strict) pushout of a cofibrant replacement (meaning cofibrant objects and maps) of the "initial" three objects to the last object is a weak equivalence, and that is the case here.

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    $\begingroup$ ACtually, Top is proper, but left properness is not due to the fact that every object of Top is cofibrant, which is false (unless you mean some goog subcategory). Moreover, it is right proper because any object is fibrant. $\endgroup$
    – Edoardo Lanari
    Commented Oct 13, 2014 at 12:25
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(I'll assume that in a general model category $\mathcal{C}$, $\mathrm{Cyl}(X)$ really means: a factorization of $A\to X$ into a cofibration $A\to \mathrm{Cyl}(X)$ followed by a trivial fibration $\mathrm{Cyl}(X)\to X$.)

A sufficient condition on objects for the map in question 1 to be a weak equivalence, is that the objects $X,Y,A$ be cofibrant. (The fact you want to use is the statement due to Reedy (which you can find at the start of the chapter on Proper Model Categories in Hirschhorn's book), that a pushout of a weak equivalence between cofibrant objects along a cofibration is a weak equivalence.)

A sufficient condition on $\mathcal{C}$ for the map in question 1 to be a weak equivalence, is that the model category be left proper.

It's an interesting fact that in Top, the this also works if $\mathrm{Cyl}(X)$ denotes the "classical" mapping cylinder construction ($X$ union a cylinder on $A$), which isn't necessarily a cofibration in the Quillen model structure. The slickest proof is to use the "excisive triad theorem", as proved by May in A Consise Course in Algebraic Topology, p.79. This also leads to a proof that Top is left proper.

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  • $\begingroup$ That is interesting. Thanks for the answer and clarification of my question. $\endgroup$
    – Joey Hirsh
    Commented Jan 8, 2010 at 20:36

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