Let $f: X\rightarrow Y$ be a continuous map of Hausdorff space and $K\subseteq X$ be a compact subset. Suppose that
$(a)$ $f|_K: K\rightarrow f(K)$ be a homeomorphism, and
$(b)$ for every $x\in K$ there exists open neighborhood $U_x$ of $x$ and $V_x$ of $f(x)$ such that $f$ stricts to a homeomorphism $U_x\rightarrow V_x$ given by $x\mapsto f(x)$ i.e. $V_x=f(U_x)$.
Then I want to prove that there exists an open subset $U\subseteq X$ containing $K$ and an open subset $V\subseteq Y$, such that $f$ restricts to a homeomorphism $U\rightarrow V$ given by $x\mapsto f(x)$.
I have no idea how to construct this open set, maybe it should be the intersection of the open neighborhood of $x\in K$? But I think it is not necessary to keep it open. Can someone help me?
Then I tried like below:
Since $\{U_x|x\in K\}$ is an open cover of $K$, then by compactness of $K$, it should have a finite subcover say $\{U_{x_i}|i=1,…,n\}$
Then define $U=\bigcup_{i=1}^n U_{x_i}$.
I can see that the restriction is continuous and surjective, but I can not show that it is also injective, and therefore the continuity of its inverse map.
Best Answer
Under the additional assumption on $Y$ that for each nbhd of a point there exists a smaller nbhd whose closure is also contained in the first nbhd (this is sometimes called T3), I have a proof.
Using the above assumption, you can find a finite cover $U_i$ of $K$ such that $f_{\bar{U_i}}$ is a closed embedding. Then you can define $V_1 = f^{-1}(f(U_1)\backslash f(K\backslash U_1))$. Note that
Now we can proceed inductively by replacing $K$ in the previous by $\bar{V_1}\cup K$, and then $\{V_i\}$ will be an open cover of $K$ (by 2. and 3.) on whose union $f$ is injective, and since $\cup V_i \subset\cup U_i$ an open map, and obviously surjective on its image.