Prove that $f$ is a homeomorphism iff $f[\overline A] = \overline {f[A]}$ Prove that $f$ is a homeomorphism iff $f[\overline A]  = \overline {f[A]}$, I know how to prove that $f$ is continuous iff $f[\overline A]  \subset \overline {f[A]}$, but how can I complete?
 A: If $f:X\to Y$ is homeomorphism, then $f(\overline{A})\subset \overline{f(A)}$ by continuity of $f$. Say $g$ is an inverse of $f$. Then using continuity of $g$, $\overline{f(A)}=fg(\overline{f(A)})\subset f(\overline{gf(A)})=f(\overline{A})$
The other direction requires $f$ to be bijection. Indeed, if this assumption is not included, we have the following example:
Take $X=\mathbb{N}$ equipped with indiscrete topology, and take $Y=\mathbb{S}=\{0,1\}$, Sierpinski space equipped with topology $\{\emptyset,\{1\},\mathbb{S}\}$. Now define $f$ by
$$f(n)=0 \;\;\forall n$$
For any nonempty $A\subset \mathbb{N}$, $\overline{A}=\mathbb{N}$, and clearly $\{0\}$ is closed in $\mathbb{S}$, so that
$$f(\overline{A})=f(\mathbb{N})=\{0\}=\overline{\{0\}}=\overline{f(A)}$$
If you do have an assumption that $f$ is bijection, then the rest is easy.
A: $f$ is continuous iff $\forall A \subseteq X: f[\overline{A}] \subseteq \overline{f[A]}$
$f$ is closed iff $\forall A \subseteq X: \overline{f[A]} \subseteq f[\overline {A}]$ (if $f$ is closed, for every $A\subseteq X$: $f[\overline{A}]$ is closed and contains $f[A]$ hence also $\overline{f[A]}$, and if the inclusion holds for all $A$, and $C \subseteq X$ is closed, then $f[C] \subseteq \overline{f[C]} \subseteq f[\overline{C}] = f[C]$, so $f[C]$ is closed and $f$ is a closed map)
So if $f$ is a bijection it's a homeomorphism iff $\forall A \subseteq X: f[\overline{A}] = \overline{f[A]}$ is a direct consequence.
