Injectivity of a locally strictly expanding map on a compact space

Question

Prove that any locally strictly expanding map on an infinite compact metric space is non-injective.

Answer 1

I have a suspicion that this is a homework question, because it appears as Lemma 13.1.3 in the book Brucks, Bruin - Topics from One-Dimensional Dynamics. However, the definition of a "locally strictly expanding map" there is different from the one suggested by YCor in the comments. Hence, I will modify the proof from the aforementioned book to show the following:

Let $f$ be a map on a compact metric space $X$ such that for every $x\in X$ there is an open neighborhood $U_x$ of $x$ such that $d(f(u),f(v))>d(u,v)$, for every $u,v\in U_x$. If additionally $f$ is injective, then $X$ is finite.

First, since $f$ is a continuous injection from a compact metric space, it is a homeomorphism onto its image $Y=f(X)$. Hence, $\{f(U_x), x\in X\}$ is an open cover of $Y$. Since $Y$ is compact, by Lebesgue theorem there is $\delta>0$ such that for every $y\in Y$ there is $x\in X$ such that $B(y,\delta)\subset f(U_x)$. In particular, if $d(y,z)<\delta$, then $d(g(y),g(z))<d(y,z)$, where $y,z\in Y$ and $g:Y\to X$ is the inverse of $f$.

It is enough to show that $Y$ is finite, since $g$ is a surjection. From compactness, $Y$ can be covered by $N$ sets of diameter at most $\frac{\delta}{2}$. We will show that $Y$ is finite by showing inductivly that it can be covered by $N$ sets of diameter at most $\frac{\delta}{2^n}$, for every $n\in\mathbb{N}$. The case $n=1$ is done. Assume that $Y=U_1\cup...\cup U_N$, where $U_i$ are of diameter at most $\frac{\delta}{2^n}$.

The function $K(x,y)=\frac{d(g(y),g(z))}{d(y,z)}$ is continuous on the compact set $\{(y,z),\frac{\delta}{2^{n+1}}\le d(y,z) \le \frac{\delta}{2^n}\}$, from where there is $\alpha\in(0,1)$ such that $\frac{\delta}{2^{n+1}}\le d(y,z) \le \frac{\delta}{2^n}$ $\Rightarrow$ $d(g(y),g(z))\le \alpha d(y,z)$. Therefore, $ d(y,z) \le \frac{\delta}{2^n}$ $\Rightarrow$ $d(g(y),g(z))\le \max\{\frac{\delta}{2^{n+1}},\alpha d(y,z)\}$, and so $diam(g(U_i))\le \max\{\frac{\delta}{2^{n+1}},\alpha diam(U_i)\}$. Since $g$ is a surjection, and $\{U_i\}$ cover $Y$, it follows that $\{V^1_i\}$ cover $Y$, where $V^1_i=g(U_i)\cap Y$.

Define also $V^{k+1}_i=g(V^k_i)\cap Y$. Applying the same argument we get that $\{V^k_i\}$ cover $Y$, and $diam(V^k_i)\le \max\{\frac{\delta}{2^{n+1}},\alpha^k diam(U_i)\}$. If $m$ is such that $\alpha^m\le \frac{1}{2}$, we get that $\{V^m_i\}$ cover $X$, and $diam(V^m_i))\le \frac{\delta}{2^{n+1}}$. This completes the proof.

Edit: it turns out that the two definitions of locally expanding maps are equivalent, as shown in this article. However, the proof is not trivial.