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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.

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

2 Answers 2

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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.

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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. $\square$

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