A restriction of a closed map is a closed map?

Question

$X,Y$ are topological spaces, $f:X \rightarrow Y $ is a closed map. Given a subset $B \subset Y$, let $A=f^{-1}(B) \subset X$. Prove that the restriction $g=f|_{A}:A \rightarrow B$ is a closed map. To prove that $g$ is a closed map, take a closed set of $A$, say $C'=C \cap A$ where $C$ is closed in $X$. We need to prove $g(C')$ is closed in $B$. But how can I get a new form of $g(C')$? I think it's strange, maybe can prove it in other way? Is this a true proposition? Or give a counter example?

Answer 1

If $C$ is closed in $A$, we can write $C= C' \cap A$, where $C'$ is closed in $X$.

Now:$$g[C] = f[C'] \cap B\tag{*}$$

and as $f$ is a closed map, $g[C]$ is closed in $B$ and so $g: A \to B$ is closed.

To see $(\ast)$ is a matter of showing two inclusions, e.g.:

• If $y \in g[C]$, $y=g(x)$ for $x \in C$. So $x \in C'$ and $x \in A = f^{-1}[B]$. It follows that $g(x)=f(x) \in f[C']$ and also $g(x)=f(x) \in B$ and so $y \in f[C']\cap B$.

• If $y \in f[C'] \cap B$, write $y=f(x)$ for $x \in C'$. As $f(x)=y \in B$, $x \in f^{-1}[B]=A$ by definition, so $x \in C' \cap A= C$ and $g(x)=f(x)=y$ so $y \in g[C]$, as required.

Note that the same holds open maps $f$: $f$ open implies $g$ is open too.

Answer 2

If $C\subseteq X$ is closed, then letting $p \in g(C\cap A)$ we see there exists some $q \in A\cap C$ such that $g(q) = p$. Notice then $g(q) \in g(A)$ and $g(q) \in f(C)$ hence $p \in g(A)\cap f(C) = B\cap f(C)$ which is closed in $B$, hence $g(A\cap C) \subseteq B\cap f(C)$. Conversely, if $p \in B\cap f(C)$, then it is simultaneously in the inverse image of $A$ and of $C$, hence there is a $q \in A\cap C$ where $g(q) = p$, but this just implies $p \in g(A\cap C)$, proving $g(A\cap C) = B\cap f(C)$.