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?