I wanted to verify the correctness of my solution to a basic point-set topology exercise. Also, could the ($2 leftrightarrow 3$) component of the proof be made more direct and enlightening?

**Proposition.** Let $f : X rightarrow Y$ be a function from space $X$ to space $Y$. The following conditions are equivalent:

- given any open $U subseteq Y$, $f^{-1}(U)$ is open,
- given any closed $F subseteq Y$, $f^{-1}(F)$ is closed, and
- given any subset $A subseteq X$, $f(mathsf{Clos}(A)) subseteq mathsf{Clos}(f(A))$.

*Proof*.

Let $f : X rightarrow Y$ be a function from space $X$ to space $Y$.

(1 $leftrightarrow$ 2) $f^{-1}(U)$ is open for every open $U subseteq Y$ $leftrightarrow$ $f^{-1}(U)^c$ is closed for every open $U subseteq Y$ $leftrightarrow$ $f^{-1}(U^c)$ is closed for every open $U subseteq Y$ $leftrightarrow$ $f^{-1}(U^c)$ is closed for every $U^c subseteq Y$ where $U$ is an open set of $Y$ $leftrightarrow$ $f^{-1}(F)$ is closed for every closed $F subseteq Y$ since complements of opens are exactly the closed sets.

(2 $rightarrow$ 3) Suppose $f^{-1}(F)$ is closed for every closed subset $F subseteq Y$. We need to show that $f(mathsf{Clos}(A)) subseteq mathsf{Clos}(f(A))$ for every subset $A subseteq X$. Let $A subseteq X$ be an arbitrary subset of $X$ and let $x in mathsf{Clos}(A)$. It remains to be shown that $f(x) in mathsf{Clos}(f(A))$ i.e. $x in f^{-1}(mathsf{Clos}(f(A)))$. By assumption, we have that $f^{-1}(mathsf{Clos}(f(A)))$ is closed so it suffices to show that $x in mathsf{Clos}(f^{-1}(mathsf{Clos}(f(A))))$. We are done if we can show that every open neighbourhood of $x$ intersects $f^{-1}(mathsf{Clos}(f(A)))$. Let $U$ be an arbitrary open neighbourhood of $x$. Because $x in mathsf{Clos}(A)$, $U$ must contain some $a in A$. Clearly, $f(a) in f(A) subseteq mathsf{Clos}(f(A))$ which is to say that $a in f^{-1}(mathsf{Clos}(A))$.

(3 $rightarrow$ 2) Suppose $f(mathsf{Clos}(A)) subseteq mathsf{Clos}(f(A))$ for every subset $A subseteq X$. Let $F subseteq Y$ be a closed subset. We need to show that $f^{-1}(F)$ is closed i.e. $mathsf{Clos}(f^{-1}(F)) subseteq f^{-1}(F)$ which is equal to $f^{-1}(mathsf{Clos}(F))$ because $F$ is closed. Let $x in mathsf{Clos}(f^{-1}(F))$. We have by assumption that

begin{equation*}

f(mathsf{Clos}(f^{-1}(F))) subseteq mathsf{Clos}(f(f^{-1}(F))) = mathsf{Clos}(F)

end{equation*}

meaning $f(x) in mathsf{Clos}(F)$ i.e. $x in f^{-1}(mathsf{Clos}(F))$ which was what we needed.