# Topology homeomorphism iff closure equals closure of image

So, I have the following problem :

Let $X$ be a topological space. Show that a bijection $f:X \to X$ is a homeomorphism iff $f(\bar A)=\overline{f(A)}$.

And I have got an online solution here. Now, I understand all the steps of the solution.

(There are typing errors in the last two statements. They should be $\forall H\subseteq S_2$ and not $S_1$ I think)

However I don't understand how the final conclusion proves the statement. Any kind of help will be highly appreciated. Thank you.

I think you are right about the typos. The last statement implies $$f^{-1}(\overline H) = \overline{f^{-1}(H)}$$ since you can take $f$ on the other side to get $$\overline{f^{-1}(H)}\subseteq f^{-1}(\overline H)$$ and you already had the other inclusion by continuity. Then you have shown the claim for $f^{-1}$, which is the same as showing it for $f$: just set $H=f(H')$, which is something you can do for every set $H'$, since $f$ is a bijection.