Local inverses of a local homeomorphism

Question

Let $(X,d)$ be a compact metric space and $f:X\to X$ a continuous map such that there exists $\delta$, so that $$f:B(x,\delta) \to f(B(x,\delta))$$ is a homeomorphism for any $x \in X$.

Show that there is a constant $0<\lambda<\delta$ such that for any $y\in X$ there are local inverses $g_1,\cdots, g_n$ of $f$ defined on $B(y,\lambda)$ satisfying $g_i(y)=x_i$ and $\{g_i(B(y,a))\}_{i=1}^n$ are pairwise disjoint.

First I showed that $f^{-1}(y)$ is finite, $f^{-1}(y)=\{x_1,\cdots,x_n\}$. Which is not difficult due to the compactness of $X$.

Now let $d(y):= \inf_{k\neq i} d(x_k,x_i)$ and take $0<r< \min\{\delta,d(y)/2\}$ so $f: B(x_j,r)\to f(B(x_j,r))$ is homeomorphic for each $1\le j \le n$.

Since $y$ is contained in the open set $\cap_{j=1}^{n} f(B(x_j,r))$ the exist $r_y$ such taht $B(y,r_y)$ is in the intersection. Now I construct the inverses

$$g_{y_i}: B(y,r_y) \to g_{y_j}(B(y,r_y))\subset B(x_i,r)$$

Thus $g_{y_i}(B(y,r_y))$ are disjoint, because $B(x_i,r)$ are disjoint.

The suggestion given to me would be to consider a cover of $X$ by balls of the form $\{B(y, r_{y}/2)\},y\in X$, so due to the compactness there is a finite coverage, $\{B(y_j, r_{y_j}/2)\}$.

The $\lambda$ that satisfies the exercise would be given by $\lambda=1/2\min{r_{y_j}}$.

But now, how would I construct the inverse $g's$? This part is confusing me.

You might find the Lebesgue number lemma helpful: https://en.wikipedia.org/wiki/Lebesgue%27s_number_lemma — CJ Dowd, May 04 '23 at 02:15
What you are trying to prove is simply false. You are probably misstating the assumption. — Moishe Kohan, May 04 '23 at 20:36
Oh, I did not notice that you have a map from $X$ to itself. — Moishe Kohan, May 04 '23 at 21:17
@MoisheKohan Do you have an idea how to finish the exercise? — Mrcrg, May 04 '23 at 21:55
Actually, I think it is still false, but I will have to check my example. — Moishe Kohan, May 04 '23 at 21:56
@MoisheKohan You can check here. This is basically Proposition 2.1.6. But it can be considered just local homeomorphism as the only thing that was used was the existence of $b$ ($\delta_0$ in my question). I didn't understand the $g_i's$ — Mrcrg, May 04 '23 at 22:03
I will add my example later on. No, your assumptions do not imply a local homeomorphism. — Moishe Kohan, May 04 '23 at 23:47

score 1 · Answer 1 · answered May 05 '23 at 00:40

What you wrote is a garbled version of the following standard fact:

Theorem. Let $X, Y$ be compact Hausdorff spaces, $f: X\to Y$ a surjective local homeomorphism. Then $f$ is a finite-to-one covering map.

See for instance here.

Now, some examples. First of all, your map $f$ may fail to be surjective. Then at each $y\in X\setminus f(X)$ there is no "local inverse" to $f$. To get such an example, consider for instance $$ X=[0,1]\cup \{2\} $$ and the map $f(x)=x, x\in [0,1]$, $f(2)=1$. Yes, this map is not a local homeomorphism, but all you required is that $f: B(x,\delta)\to f(B(x,\delta))$ is a homeomorphism for every $x\in X$ and some $\delta>0$.

Now, let's find an example of a surjective map. I will take $X$ which is the 2-point compactification (adding $\pm \infty$) of $$ \bigcup_{n\in {\mathbb Z}} [2n, 2n+1]. $$ Consider the map $f: X\to {\mathbb R}\cup \{\pm \infty\}$ which is the identity except for $$ f(x)=x+1, x\in [-2,-1]. $$ It is easy to see that $f(X)=Y$ is homeomorphic to $X$. The space $X$ is obviously metrizable, hence, take any metric on $X$ metrizing this space. Now, at the point $y=0$ the map $f$ does not admit a local continuous (I assume you are interested in continuous maps) inverse.

Local inverses of a local homeomorphism

1 Answers1