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?
Best Answer
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).