# Local homeo is not a closed map

Let $$f:M\rightarrow N$$ be a local homeomorphism with $$f^{-1}(y)$$ being infinite (I am willing to assume $$M$$ and $$N$$ are open subsets of $$\mathbb{R}^m$$ and $$\mathbb{R}^n$$). I need to show $$f$$ is not closed.

Let $$f^{-1}(y)=\{x_i | i\in I\}$$. For every $$x_i$$, there is an open neighborhood $$U_i$$ such that $$f|_{U_i}$$ is a homeomorphism. Because this function has an inverse, no other $$x_j$$ lies in $$U_i$$.

Because $$f$$ is locally continuous it must be globally continuous and $$f^{-1}(y)$$ is closed in $$M$$. The $$U_i$$ are an open cover of this closed set which admits no finite subcover (the only open set containing $$x_i$$ is $$U_i$$).

I am not sure how any of this would help me create a closed set whose image under $$f$$ is not closed.

• This seems wrong. The identity will always be a local homeo and also a closed map. Is $f$ a particular function? Jul 28, 2023 at 14:03
• @Randall, but there is no $y$ such that $f^{-1}(y)$ is infinite… Jul 28, 2023 at 14:04
• Ah, you're right. Missed that part. Jul 28, 2023 at 14:05

Start from a toy example, the standard covering $$f \colon \mathbb{R}\to \mathbb{S}^1$$. Take a small open $$U\ni 1$$ and look at the $$f^{-1}(U)=\sqcup V_i$$. After a modification we can assume that $$f$$ is a projection $$pr\colon \sqcup U_i \to U$$ with $$U_i=U$$ for all $$i$$. Pick a sequence $$x_i \in U_i$$ so that $$pr(x_i)\neq y$$ for all $$i$$ but $$\lim pr(x_i) = y$$.
Then $$f$$ is not closed because $$\{x_i: i\in \mathbb{Z}$$} is closed in $$\mathbb{R}$$ but its image is not.
Note that we didn't use much what was the domain and codomain. Except that the map need not be a covering but that is feasible by picking the sequence $$x_i$$ more carefully.
This is not true in general. Let $$X$$ be a discrete infinite set and $$f:X\to \{0\}$$ be constant. Then $$f$$ is a local homeomorphism, is closed, and $$f^{-1}(0)$$ is infinite.
Theorem. If $$f:X\to Y$$ is a local homeomorphism, $$X$$ a separable metric space with no isolated points, $$Y$$ be Hausdorff and first countable, and $$y$$ is such that $$f^{-1}(y)$$ is infinite, then $$f$$ is not closed.
Proof: First notice that $$f^{-1}(y)$$ is discrete since if $$x_n\in f^{-1}(y)$$ converges to $$x\in f^{-1}(y)$$, then taking open $$U$$ such that $$f\restriction_U$$ is a homeomorphism, we'd have $$x_n\in U$$ for large enough $$n$$. But this would imply $$x_n = x$$ for large enough $$n$$ since $$f$$ is injective on $$U$$ and $$f(x) = f(x_n)$$. Since $$X$$ is a separable metric space, it has countable spread i.e. any discrete subset of $$X$$ needs to be at most countable. Thus we can write $$f^{-1}(y) = \{x_n : n\in\mathbb{N}\}$$ where $$x_n$$ are distinct. Pick some $$r_n > 0$$ such that $$B(x_n, r_n)$$ are pairwise disjoint (this can be done by induction for example) and restriction of $$f$$ to those balls is a homeomorphism. Then pick some $$z_n\in (B(x_n, r_n)\cap f^{-1}[U_n])\setminus \{x_n\}$$ where $$U_n$$ is a local basis of $$y$$ to ensure that $$f(z_n)\to y$$ (this works because there's no isolated points in $$X$$) and $$Z = \{z_n : n\in\mathbb{N}\}$$. If $$z\in\overline{Z}\setminus Z$$, there is a subsequence $$z_{n_k}$$ convering to $$z$$. Then $$f(z) = y$$ thus $$z = x_m$$ for some $$m$$. But this is impossible as then $$z_{n_k}\in B(x_m, r_m)\cap B(x_{n_k}, r_{n_k})$$ for large enough $$k$$, but those balls were assumed to be disjoint. Thus $$Z$$ is closed. But $$y\in \overline{f[Z]}\setminus f[Z]$$ so $$f[Z]$$ is not closed. $$\square$$