Let $X, Y, Z$ be spaces, $q : X \to Y$ a quotient map, and $p : X \to Z $ a surjective continuous map. Suppose $q(x)=q(x')$ iff $p(x)=p(x')$, for all $x, x' \in X$. I hope that $p$ is also a quotient map and that $Y, Z$ are homeomorphic, but I do I have to show this?
I know that $p$ induces a continuous bijection $f : Y \to Z$ such that $f \circ q=p$, but I can't show that $f$ is a homeomorphism
Unfortunately. Consider $X=Y=Z$ as sets. Let $p=q=\text{id}$ but $X=Y$ both are discrete while $Z$ is not. Note how much stronger conditions these maps satisfy: $q$ is a homeomorphism, $p$ is a bijection and $p(x)=q(x)$.