A surjective, continuous and closed map is a quotient map

Question

I want to prove that a surjective, continuous and closed map is a quotient map. I know that a surjective map $p: X \longrightarrow Y$ is a quotient map if every set $V \subset Y$ is open $\Longleftrightarrow$ $p^{-1}(V)\subset X$ is open. Please note that $p^{-1}$ is the pre-image

Since we have continuity, one only has to prove the implication $\Longleftarrow$

Let $V\subset Y$ so that $p^{-1}(V)\subset X$ is open. $X\backslash p^{-1}(V)$ is closed and since $p$ is a closed map, $p(X\backslash p^{-1}(V))$ is closed. Here is where I get confused. If I knew that $p$ was injective, then I could keep going like this: $$p(X\backslash p^{-1}(V))=p(X)\backslash p(p^{-1}(V))$$ and since $p$ is surjective: $$p(X)\backslash p(p^{-1}(V))=Y\backslash V$$ and then we'd be done, because $Y\backslash V$ is closed and therefore $V$ is open. However, as I don't know if $p$ is injective I can't follow this path. Could someone help me?

Answer 1

Since $p$ is surjective, you know that $pp^{-1}(A) = A$ for any $A \subset Y$. Thus

$$ V^c = p(p^{-1}(V^c)) = p(p^{-1}(V)^c) $$

is closed and we are done.

Answer 2

$p: X \to Y$ is a quotient map iff

$C \subseteq Y$ is closed iff $p^{-1}[C]$ is closed in $X$ $(a)$.

(a reformulation in terms of closed sets; follows from the standard definition as $X\setminus p^{-1}[O] = p^{-1}[Y\setminus O]$ for any $O \subseteq Y$ and any map $p$)

So if $p$ is continuous and onto, we can show $\Rightarrow$ in $(a)$ by noting that $p$ is continuous and $\Leftarrow$ by noting that $p[p^{-1}[C]] = C$ as $p$ is onto and closed as $p$ is closed. So $(a)$ holds for $p$ and so $p$ is quotient.