Reasoning for definition of locally homeomorphic

Question

I am reviewing some fundamental topological concepts, and tried to recollect what it meant for a topological space $X$ to be locally homeomorphic to another topological space $Y$.

'My Definition': I would have said that $X$ is locally homeomorphic to $Y$ if for every point $x\in X$ there exists an open neighborhood $U$ of $x$ which is homeomorphic to some subspace $V\subseteq Y$.

However, when checking against what the internet said, I found that a local homeomorphism is a map $f:X\longrightarrow Y$ such that each $x\in X$ has an open neighborhood $U$ such that $f(U)$ is open in $Y$ and $f$ restricted to $U$ is a homeomorphism. I suppose saying that $X$ and $Y$ are locally homeomorphic is supposed to mean that there exists such a local homeomorphism $X\longrightarrow Y$.

1. I would like to understand whether there is a difference between my definition and the second definition I gave.

2. Why, in the second definition, does one require $f(U)$ to be open in $Y$? What could go wrong if one wouldn't require this?

Answer 1

I'm going to take a stab at this:

Your confusion comes from the fact that the modifier "locally" doesn't distribute across definitions.

Okay, that's a pretty fuzzy statement, so let me explain what I mean.

When we take a property $P$ and say something is "locally $P$", we usually mean that every point has a neighborhood with property $P$. Thus, when we say "a manifold is locally homeomorphic to $\mathbb{R}^n$", we mean every point has a neighborhood homeomorphic to $\mathbb{R}^n$. When we call a map from $X$ to $Y$ a local homeomorphism, we mean every point in $x \in X$ has a neighborhood that is mapped homeomorphically to a neighborhood of $f(x)$.

But here's the thing - under these definitions, locally homeomorphic is not equivalent to the existence of a local homeomorphism. For example, the circle is locally homeomorphic to $\mathbb{R}$, but there is no local homeomorphism from the circle to the real line - for any such map, at some point the path needs to "bend back" on itself, and near the bend point the map won't be injective. What we see here is that, though each point has a neighborhood that can be mapped homeomorphically into $\mathbb{R}$, we cannot achieve this for all points simultaneously in a single map.

This is what I mean when I say the modifier "locally" doesn't distribute across definitions: Two spaces are homeomorphic iff there exists a homeomorphism between them, but it's NOT true that two spaces are locally homeomorphic iff there exists a local homeomorphism between them.

As to your second question, I think the reason we require $f(U)$ to be open is because otherwise we aren't forcing the neighborhood of the point to match the neighborhood of the image. For example, let $X$ be a single point, and let $Y$ be the real line. We can map that point into the real line and it will be homeomorphic with its image, but the local topology around that point and its image are very different. Requiring $f(U)$ to be open prevents this map from being a local homeomorphism.

Answer 2

Let $X=\Bbb R\times\{0\}\cup \{-1,1\}\times \Bbb R$ and $Y=\Bbb R\times\{0\}\cup \{0\}\times \Bbb R$.

Then $X$ and $Y$ are atureal-locally homeomorphic (in fact, in both directions): $(\pm1,0)\in X$ has cross-shaped neighbourhoods, just like $(0,0)\in Y$, all other points have interval-shaped neighbourhoods.

But there does no exist a local homeomorphism $f\colon X\to Y$. Such $f$ would have to map $(\pm1,0)\mapsto (0,0)$. On the compact interval $[-1,1]\times\{0\}$, $x\mapsto \|f(x)\|$ assumes a maximum and therefore the image of $(-1,1)\times\{0\}$ is not open in $Y$.