Let $f:X\to Y$ and $g: Y\to Z$ surjective, continuous functions. Proof or disprove, that if $g\circ f$ and $g$ are quotient maps, then $f$ is a quotient map.
I think this is not correct, and I want to give a counterexample. Suppose $Z=\{\ast\}$ and $f$ is a quotient map.
Let $\emptyset\neq U\subseteq Y$ and $\emptyset\neq f^{-1}(U)\subset X$ be NOT open. Then $g\circ f(f^{-1}(U))=g(U)\subseteq Z$. Hence $g(U)=\emptyset$ or $g(U)=Z$. Since $U\neq\emptyset$ it is $g(U)=Z$ which is open in $Z$. Since $g$ is continuous it is $U\subseteq Y$ open. Contradiction, because $f^{-1}(U)$ was NOT open.
Is this correct? Thanks in advance.
Perhaps a simpler argument following your idea would be better. For any surjective continuous map $f$, let $Z$ be the point, then $g$ is uniquely defined, and $g\circ f$ and $g$ are automatically quotient maps. Thus if the statement were true, all surjective, continuous maps would be quotient maps, and this is easily seen to be false.
One easy example of such a map is the identity mapping from the discrete topology on two points to the trivial topology on two points. This isn't a quotient map because the preimage of a singleton set is open in the discrete topology, but singletons aren't open in the trivial topology.