# Continuous map of Hausdorff space

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.

• Arbitrary intersections of open sets need not be open... Jul 25, 2021 at 21:44
• @Henno Brandsma, yes I know this… so I don’t know how to construct the open set I need… Jul 25, 2021 at 22:14
• Is there a problem if we just take $U=\bigcup_{x \in K} U_x$? Try to find the potential obstruction... Jul 25, 2021 at 22:37
• @HennoBrandsma, I think the obstruction might be that, I can not prove it is a bijection between the $U=\bigcup_{x\in K}U_x$ and $V$... Jul 26, 2021 at 15:18

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

1. $$V_1\subset U_1$$
2. $$V_1$$ is open
3. by injectivity on $$K$$ that $$K\cap V_1=K\cap U_1$$
4. $$f_{|\bar{V_1}\cup K}$$ is a closed topological embedding

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.

• If you want, I can add some more details. Aug 4, 2021 at 11:06

I think your choice $$U = \bigcup_{i=1}^nU_{x_i}$$ might be problematic.

Assume that we have $$K \subsetneq U$$, i.e. $$U \backslash K \neq \emptyset$$, and assume that $$\#(U \backslash K) \geq 2$$. Let $$Y := X/\sim$$, where $$x \sim x' :\iff x,x' \in U\backslash K$$, and let $$f : X \twoheadrightarrow X/\sim$$ be the canonical projection.

Then, $$f|_U : U \to V=f(U)$$ is injective, if for $$x,x' \in U$$, with $$x \neq x'$$, we have $$f|_U(x) \neq f|_U(x')$$. However, if we choose $$x,x' \in U \backslash K$$, such that $$x \neq x'$$ (here, we used that $$\#(U\backslash K)\geq 2$$), then we always have $$f|_U(x) = f|_U(x')$$, because $$x \sim x'$$.

This construction doesn't contradict condition (a), since $$\sim$$ doesn't affect the elements of $$K$$. I think it doesn't contradict condition (b), if you choose $$X$$ to be a "nice" space, like some subspace of $$\mathbb{R}^d$$ for instance.

I'm not entirely sure this is correct. Am I missing something?

• Thank you for your answer, I think you are right. So I can only go back to the beginning to think about how to construct the $U$ I need... Jul 26, 2021 at 14:20