# Homotopy pushouts and induced maps

Suppose we are in a proper closed model category and consider a commutative square $$\begin{array}{rcl} A&\to& B\\ \downarrow&&\downarrow\\ C&\to&D \end{array}$$ in its homotopy category. This square is canonically isomorphic in the homotopy category to a square in which all morphisms are morphisms of the model category (we assume here to have (fixed) functorial factorizations such that all this can be done nicely). Hence, w.l.o.g. the square above is a commutative square in the model category.

If we (functorially) factorize the morphism $A\to B$ into a cofibration followed by a trivial fibration like $A\to B'\to B$, we can form the categorical pushout $P$ of $$\begin{array}{rcl} A&\to& B'\\ \downarrow&&\\ C& \end{array}$$ There is an induced map $i:P\to D$ in the homotopy category and one can define the initial square to be a homotpy pushout square if the map $i$ is an isomorphism in the homotopy category.

If one considers now such a homotopy pushout square $$\begin{array}{rcl} A&\to& B\\ \downarrow&&\downarrow\\ C&\to&D \end{array}$$ in the homotopy category with morphisms $B\to E$ and $C\to E$ such that everything commutes, there should be a (unique?) morphism $e:D\to E$ in the homotopy category obtained by the construction above.

My question is: Why isn't this construction the colimit in the homotopy category? I have heard that homotopy colimits are not categorical colimits in the homotopy category. Does it fail with the uniqueness of the morphism $e$? The existence of $e$ should be sure. Does one have at least the existence of an induced morphism (in the homotopy category!) for general homotopy (co)limits?

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## 2 Answers

I think you don't even have the existence of $e$.

The problem is:

1. What are the morphisms in the homotopy category, and
2. When two morphisms are equal in the homotopy category.

As for the first question, in general the answer is that a morphism from $A$ to $B$ in the homotopy category is a sequence of maps of the original category

$$A= A_0 \stackrel{\sim }{\longleftarrow} A_1 \longrightarrow A_2 \stackrel{\sim}{\longleftarrow} \dots \stackrel{\sim}{\longleftarrow} A_n \longrightarrow B$$

where the ones going to the left are weak equivalences.

In general, it's difficult to tell when two such sequences are equal. The best think you can have is that your class of weak equivalences admits a (right, left) calculus of fractions. Then, arrows in the homotopy category are sequences of length $2$

$$A \stackrel{\sim}{\longleftarrow } A_1 \longrightarrow B$$

and equality between two such sequences means that some diagram made up from four triangles is commutative. In general, equality involves bigger commutative diagrams in the original category. For more information, see Dwyer, Hirschhorn, Kan, Smith, "Homotopy limit functors on model categories and homotopical categories", AMS Math. Surveys and Monographs, 113.

So, try now to say what's the meaning for two such arrows (sequences) starting from your $B$ and $C$ and ending in $E$ to form a commutative diagram in the homotopy category. Then, try to imagine how this could imply the existence of some $e: D \longrightarrow E$.

EDIT. Maybe I can add another point of view. If $\cal{C}$ is a model category, then its (Quillen) homotopy category (obtained inverting its weak equivalences) $\mathrm{Ho} \cal{C}$ is equivalent to the (classical) homotopy category $\pi_{cf}(\cal {C})$; i.e., the category whose objects are the fibrant-cofibrant ones and maps are homotopy classes of maps of $\cal{C}$.

So, this $\pi_{cf}(\cal {C})$ allows you a more specific representations of maps of the (equivalent, Quillen) homotopy category: here, they are "real" (homotopy classes of) maps of $\cal{C}$, not unmanageable sequences.

Even in this equivalent homotopy category, the problem remains: when you have two (classes) of maps $B \longrightarrow E$ and $C \longrightarrow E$, making your diagram commutative in $\pi_{cf}(\cal {C})$... "Commutative" in $\pi_{cf}(\cal {C})$ means commutative up to homotopy in $\cal{C}$. And it doesn't matter if your objects are fibrant-cofibrant and your morphisms cofibrations: from the universal property of the push-out in $\cal{C}$ you cannot deduce the existence of $e: D \longrightarrow E$ if you have commutativity up to homotopy in the diagram involving $B \longrightarrow E$ and $C \longrightarrow E$.

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I'm sorry. I'm not familiar with those "homotopy push-out". – a.r. Oct 5 '11 at 17:20
Ncatlab (ncatlab.org/nlab/show/homotopy+pullback) says that these notations of "global" and "local" homotopy pushouts coincide, but I don't see why. – John Zhao Oct 5 '11 at 18:04

It's not true that the homotopy colimit is the colimit in the homotopy category. In fact, colimits (other than disjoint unions, say) tend not to exist there.

Here's a bit of intuition for homotopy colimits. (A nice reference is Dan Dugger's primer on the subject.) Suppose you have a diagram of, say spaces or simplicial sets, as in $$\begin{array}{rcl} A&\to& B\\ \downarrow&&\downarrow\\ C&\to& \end{array}$$

To form the push-out in an ordinary sense, you would take the disjoint union $B \sqcup C$, and then quotient by the relation that for each $a \in A$, the image of $a$ in $B$ is identified with that in $C$. This, however, is not homotopy invariant, because the operation of quotienting in this way is not well-behaved homotopically. A better variant (from this point of view, at least) is not to quotient by this relation, but to glue in paths for it. That is, if $i: A \to B, j: A \to C$, then to form the homotopy push-out of this diagram, you should draw in paths from $i(a)$ to $j(a)$ for each $a \in A$. This is one explicit construction of the homotopy push-out which works for spaces that are nice (say, CW complexes) or for arbitrary simplicial sets (for the more general theory, you should look up the Bousfield-Kan formula). You can check that in this case the operation just described is homotopically well-behaved, and that it is weakly equivalent to the usual colimit for projectively cofibrant diagrams.

So, let's take this as our explicit construction of the homotopy push-out of spaces or simplicial sets. Now, we want to describe maps from the homotopy push-out, which I'll call $D$, into an arbitrary space $X$. By definition, this is given by maps $B \to X$, $C \to X$, and a homotopy $A \times I \to D$ of the restriction $A \to B \to X$ with $A \to C \to X$. This follows from the other construction as $D$ by gluing in lots of paths. Note that the homotopy itself is part of the data; it's not enough to just give two maps that happen to be homotopic. We can think of this as a space of maps, and its connected components are the morphisms $[D, X]$ in the homotopy category.

Now, if the space of maps $D \to X$ were just the space of maps $B \to X, C \to X$ whose restriction to $A$ were homotopic, then you're right: $D$ would be the push-out in the homotopy category. But the problem is that the homotopy has to be included in the data. Here's an explicit example to illustrate this. Let $B = C = \ast$. Then the homotopy push-out is given by drawing paths $\ast \to \ast$ for each $a \in A$; this, in other words, is the suspension of $A$ (whether it's reduced or unreduced depends on whether you work in the pointed or unpointed category). What the above says is that to give a map $\Sigma A \to D$ is to give two points of $D$, and a homotopy between the two constant maps $A \to D$ given by these points; this is reasonable. More to the point, in this case the homotopy push-out is definitely not the push-out in the homotopy category, because $\ast$ is final in the homotopy category and that push-out would be $\ast$.

Finally, note that even homotopy inverse limits are not inverse limits in the homotopy category (e.g. because of the Milnor exact sequences with $\lim^1$ terms).

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