let $hTop\_\*$ denote the homotopy category of pointed spaces. I believe that it has no pushouts, in general. the reason is that you can't expect the involved homotopies to be compatible. can anyone give an explicit example, with proof? I know that homotopy colimits are related to this, but they don't seem to be categorical colimits, so I don't think that they fit here.


especially I'm interested in the following special case: let $G= \langle X | R \rangle$ a presentation of a group and consider the resulting map $\omega : \vee\_{r \in R} S^1 \to \vee\_{x \in X} S^1$. does the cokernel of $\omega$ exist in $hTop\_\*$? ~~in $Top\_\*$, the cokernel is just~~  consider the 2-dimensional CW-complex $Q$, which is optained from $\vee\_{x \in X} S^1$ via the attaching map $\omega$. now if $f : \vee\_{x \in X} S^1 \to T$ is a pointed map such that $f \omega$ is nullhomotopic, it is easy to see that it extends to a map $\overline{f} : Q \to T$. but I think that we cannot expect that $\overline{f}, \overline{g}$ are homotopic, when $f,g$ are homotopic: the homotopies between $f$ and $g$ don't have to be compatible. can you give an example for that? probably it already works for $\omega : S^1 \to S^1, z \mapsto z^2$, thus $Q = \mathbb{R} P^2$.


anyway, this only would show that $Q$ is not the cokernel in the category $hTop\_\*$. the proof, that the cokernel does not exist at all, will be even more difficult and I don't know how to approach it. 


you may also replace the category by $hCW\_\*$ (CW-complexes), $hCG\_\*$ (compacty generated spaces) etc., if it's useful.



categorical homotopy colimits
Your example (the "cokernel" of the multiplication by 2 map) also works.


Consider the diagram $S^1 \leftarrow S^1 \rightarrow D^2$ in the based homotopy category of CW-complexes, where the left-hand map is multiplication by 2. Suppose it had a pushout $X$ in the homotopy category. Then for any $Y$, $[X,Y]$ is isomorphic to the set of 2-torsion elements in $\pi\_1(Y)$.


Taking $Y = S^0$, we find $X$ is connected.


Taking $Y = K(\pi,1)$, we find that $\pi\_1(X)$ must be isomorphic to $\mathbb{Z}/2$. This means that there is a map from ${\mathbb{RP}^2}$ to $X$ inducing an isomorphism on $\pi\_1$, and that there is a map $X \to K(\mathbb{Z}/2,1)$ that also induces an isomorphism on $\pi\_1$.


Net result, we get a composite sequence of maps $\mathbb{RP}^2 \to X \to \mathbb{RP}^\infty \to \mathbb{CP}^\infty$. The final space is simply connected, so the map from $X$ would be nullhomotopic and hence so would the map from $\mathbb{RP}^2$.


However, the composite of the first two maps is an isomorphism on $\pi\_1$, hence on $H\_1$. Looking at induced maps on the second cohomology group $H^2$, we get the sequence of maps:
$$\mathbb{Z}/2 \leftarrow H^2(X) \leftarrow \mathbb{Z}/2 \leftarrow \mathbb{Z}$$
The rightmost map is surjective, the composite of the two leftmost maps is an isomorphism by the universal coefficient theorem, and the composite of the two rightmost maps is supposed to be nullhomotopic and hence zero. Contradiction.


