Pre-images and local homeomorphisms 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.
 A: 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.
A: 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.
