If a continuous surjection $f$ is open/closed, then it is an identification.

Question

Let $f: X \rightarrow Y$ be a continuous surjection between topological spaces. I have to show that: If f is open/closed, then it is an identification.

Definition of an identification: A continuous surjection $f: X \rightarrow Y$ between topological spaces is an identification if $f^{-1}(V) \subset X$ is open implies that $V \subset Y$ is open.

I first tried to prove that $f$ is an identification if $f$ is open.

$f$ is open means that for every open $U \subset X$, we have that $f(U) \subset Y$ is open.

So assume that $f^{-1}(V) \subset X$ is open and let $V \subset Y$. Now we have to prove that $V \subset Y$ is open. Since $f$ is open, $f(f^{-1}(V)) \subset Y$ is open. But I don't know if $f$ is bijective. So I can't assume that $f(f^{-1}(V)) = V$ I think. But I don't know how to prove this statement otherwise.

The other statement: f is an identification if f is closed. And $f$ is closed if for every closed $C \subset X$, we have that $f(C) \subset Y$ is closed. I think I have to do something similar but now with the complement of $C$ in $X$ (since the complement is open in $X$).

Answer 1

For a surjection $f: X \to Y$ we have that $f[f^{-1}[A]] = A$ for all $A \subseteq Y$: if $a \in A$ we have $x \in X$ so that $f(x)=a$, using surjectivity, and then by definition $x \in f^{-1}[A]$ (as $f(x)=a \in A$) and so $a= f(x) \in f[f^{-1}[A]]$. And if $y \in f[f^{-1}[A]]$, there is some $x \in f^{-1}[A]$ with $f(x)=y$, and because $x \in f^{-1}[A]$ we know that $y=f(x) \in A$ (so that inclusion always holds regardless of surjectivity).

So indeed if $f^{-1}[O]$ is open and $f$ is open we know that $O=f[f^{-1}[O]]$ is the image of an open set, thus open.

And $f$ (continuous, surjective) is an identification iff for all $C \subseteq Y$ we know that if $f^{-1}[C]$ closed, then $C$ is closed. (Standard equivalence to the open variant definition, proved by observing $f^{-1}[Y\setminus A]=X\setminus f^{-1}[A]$ for all $A \subseteq Y$.)

So we can apply the same argument for $f$ closed as well.

Answer 2

You don't need it to be bijective. $f$ being surjective is enough to conclude $f(f^{-1}(V))=V$. The proof is very easy, so I'll leave it to you. Just show two sided inclusion.

If $f$ is closed then using the same argument $f^{-1}(V)\subseteq X$ being closed in $X$ implies $V$ is closed in $Y$. So now, suppose $f^{-1}(V)\subseteq X$ is open in $X$. Then its complement $X\setminus f^{-1}(V)=f^{-1}(Y\setminus V)$ is closed in $X$ by definition. Hence $Y\setminus V$ is closed in $Y$, and so $V$ is open in $Y$.