This is a question on a homework for a topology course.
Let $f:X\twoheadrightarrow Y$ and $g:Y\twoheadrightarrow Z$ be continuous surjections. Prove that if $g\circ f$ is a quotient map, then $g$ is a quotient map.
Proof. Since $g$ is a continuous surjection, to show $g$ is a quotient map it suffices to show that a set $W\subseteq Z$ is open if and only if $g^{-1}(W)$ is open in $Y$. Let $W$ be an open subset of $Z$. Since $g$ is continuous, then $g^{-1}(W)$ is open in $Y$. Conversely, let $g^{-1}(W)$ be open in $Y$, and then by the continuity of $f$ we have that $f^{-1}(g^{-1}(W))$ is open in $X$. Since $g\circ f$ is a quotient map, thus $W$ is open in $Z$. $\Box$
Is this correct?
Best Answer
It is correct.
Let me make a remark on the definition of a quotient map.
If you understand a map to be a continuous function, then a surjective map $g : Y \to Z$ is a quotient map provided $$\forall W \subset Z : g^{-1}(W) \text{ open in } Y \implies W \text{ open in } Z . \tag{1}$$ The reverse implication $W \text{ open in } Z \implies g^{-1}(W) \text{ open in } Y$ is automatically satisfied because $g$ is continuous (as you observed in your proof); it is therefore unnecessary to include it in the definition.
On the other hand, if you interpret map as a synonym for function, then you must require $$\forall W \subset Z : g^{-1}(W) \text{ open in } Y \Longleftrightarrow W \text{ open in } Z \tag{2}$$ because continuity is not automatically covered by $(1)$.