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I want to prove that if $f: M \to N$ is a local homeomorphism, then for all $y \in N$ we have $f^{-1}(\{y\}) \subset M$ closed and discrete.

Here's the catch: this is from an exercise sheet from over a year ago, so it was written sloppily and it is not clear if we can assume $M$ and $N$ to be metric spaces, or if they are topological spaces in general.

I had solved it, but reading again I think that there are some things off in my solution, so I would like some input and to know how to fix it, if needed.

What I had done: Let $y \in N$ be arbitrary. Take $x \in f^{-1}(\{y\})$. Since $f$ is a local homeomorphism, exists $U \subset M$ open containing $x$ such that $f\big|_U : U \to f(U)$ is a homeomorphism. In particular, $f$ is bijective. I claim that $\{x\} = U \cap f^{-1}(\{y\})$. If $x' \in U \cap f^{-1}(\{y\})$, we have that $f(x) = f(x') = y$, and since $x,x' \in U$ and $f$ is injective in $U$, we have that $x = x'$. So $x$ is an isolated point of $f^{-1}(\{y\})$ (more precisely, $\{x\} = U \cap f^{-1}(\{y\})$ is the intersection of an open set with $f^{-1}(\{y\})$, hence closed in $f^{-1}(\{y\})$). Since $x$ was arbitrary, every point of $f^{-1}(\{y\})$ is isolated, so $f^{-1}(\{y\})$ is discrete. And for closedness, it suffices to note that $f^{-1}(\{y\})' = \varnothing \subset f^{-1}(\{y\})$, and every set containing all of its limit points is closed.

Issues:

  • It seems I didn't actually prove that $f^{-1}(\{y\})' = \varnothing$, but only that $x \in f^{-1}(\{y\}) \implies x \not\in f^{-1}(\{y\})'$, so I would have to make an argument for the points $x \not\in f^{-1}(\{y\})$. Maybe it is trivial, but I'm not seeing it.

  • I didn't used continuity of $f\big|_U$. This bothers me. Proving continuity of $f$ is the following exercise, which I managed to do (and I'm happy with my work there).

  • "$f^{-1}(\{y\})'\subset f^{-1}(\{y\}) \implies f^{-1}(\{y\})$ closed" assumes at least $T_1$, no?

  • Do we need to assume $M$ and $N$ to be metric spaces for this to work?

Thanks.

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2 Answers 2

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Issue 0:

So $x$ is an isolated point of $f^{-1}(\{y\})$ (more precisely, $\{x\} = U \cap f^{-1}(\{y\})$ is the intersection of an open set with $f^{-1}(\{y\})$, hence closed in $f^{-1}(\{y\})$).

That may be a typo here, a writo in your old work that you copied here, or a silly mistake, but probably one of the former two: Of course $\{x\}$ being the intersection of an open set with $f^{-1}(\{y\})$ means that $\{x\}$ is open in $f^{-1}(\{y\})$. [Once the discreteness of the fibre is established, it follows that $\{x\}$ is closed in $f^{-1}(\{y\})$ too, of course.]

With that fixed, your argument that $f^{-1}(\{y\})$ is discrete in the subspace topology is correct.

Issue 1: Global homeomorphisms are local homeomorphisms, and if $f$ is a global homeomorphism, $f^{-1}(\{y\})$ is closed for all $y$ if and only if $N$ is a $T_1$-space.

So it is a necessary condition that we require $N$ to be a $T_1$-space to be able to deduce that $f^{-1}(\{y\})$ is closed for all $y$ and all local homeomorphisms $f\colon M \to N$. But since local homeomorphisms are continuous, that is also sufficient to deduce that $f^{-1}(\{y\})$ is closed.

Since the discreteness of the fibre was shown without any conditions on $M$ and $N$, the desired conclusion follows under the sole assumption that $N$ is $T_1$. Of course, a local homeomorphism $f\colon M \to N$ can then only exist if $M$ is also $T_1$.

But indeed, your argument for closedness is incorrect, you have never shown that $f^{-1}(\{y\})' = \varnothing$. However, the implication $A' \subset A \implies A = \overline{A}$ is true in all topological spaces.

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  • $\begingroup$ Thanks for the thorough answer! +1. Time for feedback. About Issue $0$, that was a typo, I meant "hence open in $f^{-1}(\{y\})$". Is there a way to show closedness of $f^{-1}(\{y\})$ without appealing directly (at least) to continuity? Because proving that every local homeomorphism is continuous came after that. From the hypothesis on $N$ being $T_1$ onwards, all ok! (maybe you got a typo there, "can only exist if $M$ is also $T_1$"?) $\endgroup$ Commented Jul 12, 2015 at 22:40
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    $\begingroup$ Oh, yes, that's a typo, thanks. As to the closedness of the fibre, suppose that $z \in \overline{f^{-1}(\{y\})}$. Take an open neighbourhood $U$ of $z$ such that $f\lvert_U \colon U \to f(U)$ is a homeomorphism (to an open subset of $N$). Assuming $N$ is $T_1$, if we had $f(z)\neq y$, we could find an open neighbourhood $V$ of $f(z)$ with $y\notin V$. Then $U \cap f^{-1}(V)$ is a neighbourhood of $z$ that doesn't intersect $f^{-1}(\{y\})$, contradicting the assumption $z \in \overline{f^{-1}(\{y\})}$. We need the local continuity, there's no way around that, as far as I can see. $\endgroup$ Commented Jul 12, 2015 at 22:58
  • $\begingroup$ Using local continuity is ok. I didn't wanted to use global continuity, only. That was very clear, thanks again! $\endgroup$ Commented Jul 12, 2015 at 23:04
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The fact is not valid for general topological spaces. For example, the identity $i:\mathbb{R} \rightarrow \mathbb{R}$, when $\mathbb{R}$ is with the antidiscrete topology (only the empty set and the whole $\mathbb{R}$ are open). It is a local homeomorphism, but the inverse image of a point is not closed (but is discrete).

The discretude of inverse image of points is valid in general: each point has an open neighborhood that makes the aplication homeomorphism, then bijection, and not two points of $f^{-1}(y)$ can lie on this neighborhood.

When the singleton $\{y\}$ is closed, then $N$ $\backslash$ $\{y\}$ is open. $f^{-1}(N$ $\backslash$ $\{y\}) = M$ $\backslash$ $f^{-1}(\{y\})$ is open and follows that $f^{-1}(\{y\})$ is closed. Note that the hypothesis that $\{y\}$ is closed is not just sufficient, but necessary when $f$ is surjective. Because $f^{-1}(\{y\})$ closed $\Rightarrow M \backslash$ $f^{-1}(\{y\})$ open, and because an local homeomorphism is an open map, $N$ $\backslash$ $\{y\}$ is open, then $\{y\}$ is closed.

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  • $\begingroup$ Nice observation about the indiscrete topology. Thanks. I just don't see how a singleton being compact implies that the fibres are closed. $\endgroup$ Commented Jul 12, 2015 at 23:00
  • $\begingroup$ You don't see because is bullshit hahaha. My mistake. $\endgroup$ Commented Jul 12, 2015 at 23:09
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    $\begingroup$ Don't worry haha $\endgroup$ Commented Jul 12, 2015 at 23:09

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