$( \mathscr{X}, d_x)$ and $( \mathscr{Y}, d_y)$ metric spaces ..Proof Verification

Question

Let $( \mathscr{X}, d_x)$ and $( \mathscr{Y}, d_y)$ be two metric spaces and

$ D \subset \mathscr{X} $ a compact set.

To show is, that if

$f: D \rightarrow \mathscr{Y} $is continous and injective, then

$ f^{-1} :f(D) \rightarrow D $

is also continous.

My Idea:

Since $D$ is compact, it is also closed. $$ (f^{-1})^{-1} (D)= f(D) $$ And since $f$ is continous and $D$ compact, is also $f(D)$ compact. As Subset of the metric space $\mathscr{Y} $ is $f(D)$ also closed. So the inverse image of a closed image is also closed, therefore is $f^{-1}$ is also continous.

Does that makes sense to you? :) but where does the injectivity comes in? Any help is appreciated !

Answer 1

Let $g=f^{-1}:f(D)\to D$. We show that $g$ is continuous if $g^{-1}(E)$ is closed for every $E\subset D$ closed.

Fix $E\subset D$ closed. Since $D$ is compact, $E$ is also compact. Therefore $g^{-1}(E)=f(E)$ is compact, hence closed.

Notice that the proof holds for every $f:X\to Y$ where $X$ is compact and $Y$ is Hausdorff. No need for a metric structure.