Homeomorphism theorem

I need to show that if $f: X \rightarrow Y$ is 1-1 and $X$ and $Y$ are metric spaces, then if $\forall A\subset X, f(\overline{A})=\overline{f(A)} $ then $f$ is homeomorphism.

1) Assume $f$ is 1-1 and $\forall A\subset X, f(\overline{A})=\overline{f(A)} $. I need to show that f is a homeomorphism.

It is sufficient to show that f is continuous and $f^{-1}$ as well is continuous. I cannot show these steps unless f is onto since I need to use $f(f^{-1}(A))=A $ often.
Please help me with this.
Thank you.

Solutions Collecting From Web of "Homeomorphism theorem"

Suppose that $f:X\to Y$ is bijective. Then there are a few useful things to note/prove:

(1) $f$ is continuous if and only if $f\left(\overline{A}\right)\subseteq\overline{f(A)}$ for all $A\subseteq X$. In the same way, $f^{-1}:Y\to X$ is continuous if and only if $f^{-1}\left(\overline{A}\right)\subseteq\overline{f^{-1}(A)}$ for all $A\subseteq Y$.

(2) $f$ is an open map if and only if $f^{-1}:Y\to X$ is continuous.

(3) Given $A\subseteq X,B\subseteq Y$, we have $f(A)\subseteq B$ if and only if $A\subseteq f^{-1}(B)$.

Is that enough to get you started?


Now, let’s suppose that $f:X\to Y$ is 1-to-1, but not necessarily onto. Putting $Z=f(X)$, we have $f:X\to Z$ is a bijection, and $f^{-1}:Z\to X$ is perfectly well-defined. Now, I’m going to give slightly altered versions of the above hints (and add another one) to fit your change to the question. First, though, I’m going to define the following notation to avoid confusion:

Given a subset $S$ of a topological space $T$, we will now denote the closure of $S$ in $T$ by $\text{cl}_T(S)$. (Since we’re dealing with several different spaces, it’s a good idea to keep track of where things are.)

Now, we want to show that $f$ is a homeomorphism from $X$ to $Z$ (rather than to $Y$) if and only if $$\forall A\subseteq X,\:f\bigl(\text{cl}_X(A)\bigr)=\text{cl}_Y\bigl(f(A)\bigr).\tag{#}$$

Now, let’s amend my list of useful notes/results-to-be-proved.

(1) $f:X\to Z$ is continuous if and only if $f\bigl(\text{cl}_X(A)\bigr)\subseteq\text{cl}_Z\bigl(f(A)\bigr)$ for all $A\subseteq X.$ In the same way, $f^{-1}:Z\to X$ is continuous if and only if $f^{-1}\bigl(\text{cl}_Z(A)\bigr)\subseteq\text{cl}_X\bigl(f(A)\bigr)$ for all $A\subseteq Z.$

(2) $f:X\to Z$ is an open map if and only if $f^{-1}:Z\to X$ is continuous.

(3) Given $A\subseteq X,B\subseteq Y$, we have $f(A)\subseteq B$ if and only if $A\subseteq f^{-1}(B)$. (This is also true if we replace $Y$ with $Z.$)

(4) $S\subseteq Z$ is closed in $Z$ if and only if there is some $F\subseteq Y$ such that $S=F\cap Z$ and $F$ is closed in $Y;$ thus, $\text{cl}_Z(S)=Z\cap\text{cl}_Y(S)$ for all $S\subseteq Z.$

Hopefully that will get you where you need to go.