Prove that the diagram $q: X \to Y$, $h \circ q^{-1}: Y \to Z$, $h: X \to Z$ commutes. Suppose that the onto map $q: X \to Y$ is an identification, and $h: X \to Z$ is continuous. Assume $h \circ q^{-1}$ is single valued. Prove:
1) The function $h \circ q^{-1}: Y \to Z$ is continuous and the diagram $q: X \to Y$, $h \circ q^{-1}: Y \to Z$, $h: X \to Z$ commutes.
2) The composition $h \circ q^{-1}$ is an open (closed) map iff $h(U)$ is open (closed) whenever $U$ is an open (closed) set such that $U = q^{-1}(q(U))$.
For 1), I think I have to show that given an open set $U \subseteq Z$ and show that $(h \circ q^{-1})(U)$ is open. So I have $q(h^{-1}(U))$ has to be open. But $h^{-1}(U)$ is open, since $h$ is continuous. So I have to show that $q$ of this open set is open. I know this is where the single valued-ness of $h \circ q^{-1}$ comes in, and that $q$ is an identification, but I can't seem to figure out how to use it. Is there anything else I have to show here to conclude that the diagram commutes?
For 2), I want to argue that since $q$ is an identification, we have that $q^{-1}(U)$ is open if $U$ is open, and closed otherwise. Hence the whole map is open iff $h(U)$ is open whenever $U$ is open. But it feels like I'm missing something, since I'm not using all the hypotheses. Where does this argument go wrong?
 A: The notation $h\circ q^{-1}$ here is weird, but we'll go with it. Given an open $U\subset Z$ we have to show, actually, than the inverse image is open, that is, $(h\circ q^{-1})^{-1}(U)$. This is $(q^{-1})^{-1}(h^{-1}(U))=q(h^{-1}(U))$. Now, since we assumed $h\circ q^{-1}$ is single-valued, $h^{-1}(U)$ is a union of equivalence classes, and open since $h$ is continuous. Identification maps are open on unions of equivalence classes, that is, on open sets $V$ such that $V=q^{-1} q(V)$, by the definition of the identification topology. Thus $(q^{-1})^{-1}(h^{-1}(U))=q(h^{-1}(U))$ is open. There's nothing nontrivial to say about the diagram commuting. 
For (2), suppose $h\circ q^{-1}$ is open. Then if $U=q^{-1}(q(U))$, $h(U)=h(q^{-1}(q(U)))=(h\circ q^{-1})(q(U))$ is open. Conversely, if $h(U)$ is open for $U=q^{-1}(q(U))$ then let $W$ be open in $Y$, so $W=q(U)$ for some open $U$ in $X$. Then $(h\circ q^{-1})(W)=(h\circ q^{-1})(q(U))=h(q^{-1}(q(U)))$ is open because certainly $q^{-1}(q(U))=q^{-1}(q(q^{-1}(q(U))))$. 
The "and closed otherwise" in your proposed argument isn't true: you mean to say $q^{-1}(U)$ is open if and only if $U$ is open, and $q^{-1}(U)$ is closed if and only if $U$ is closed. Otherwise your argument probably worked-but you can assume less than that $h$ is open. Observe also that we didn't need $h$ to be continuous in part (2).
