I am trying to construct coequalizers in the category $Top^h_\ast $ whose objects are pointed topological spaces and whose morphisms are homotopy classes of pointed maps under homotopies that preserve the base point. I know that the coequalizer of $f: C \rightarrow Y $ and $g: C \rightarrow Y $ in $Top$ is given by the quotient $Y / (f(x) \sim g(x)) $ together with its projection map $\pi$. I strongly suspect that the same object with $[\pi]$ is the coequalizer in $Top^h_\ast $, but I can't produce a rigorous proof, one of the issues being that this is not really well defined since the quotient space depends on a choice for representatives of the classes of maps of which I'm trying to construct a coequalizer (even though its homotopy type doesn't).
One textbook that I have claims that a coequalizer can be constructed as the double mapping cone, which I think is homotopy equivalent to the quotient I have described.
Best Answer
The coequalizer, in $Top^h_*,$ of the diagram $A\rightrightarrows B$ should be a space $C$ such that, for every map $a:B\to X$ whose restrictions to $A$ are homotopic, there is a unique factorization of $a$ through $C,$ up to homotopy. The problem is the condition that the restrictions to $A$ merely admit a homotopy between them--the only possible way for $C$ to contain the information that these maps are homotopic is for $C$ to be equipped with a map from $A\times I$ realizing that homotopy. But such a $C$ would represent maps from $B$ with a chosen homotopy between their two restrictions to $A,$ which is simply not the same functor.
The above is the heuristic explanation of why it should be quite clear after a bit of thought that $Top^h_*$ cannot have coequalizers in general. Actually constructing definitive counterexamples is a bit technical, but here is an example of a span admitting no pushout in $Top^h.$ You should be able to adapt it to the case at hand if you're sufficiently motivated.