4

According to Wikipedia :

A function $f:X→Y$ between two topological spaces is called a local homeomorphism if every point $x\in X$ has an open neighborhood $U$ whose image $f(U)$ is open in $Y$ and the restriction $f\rvert_U:U→f(U)$ is a homeomorphism (where the respective subspace topologies are used on $U$ and on $f(U)$.

However, if we only ask that for every $x \in X$ there is a neighborhood $U$ such that $f\rvert_U:U→f(U)$ is a homeomorphism, do we get a local homeomorphism? It seems to me that yes but I'm sure I'm wrong because this definition would be simpler.

Let $x \in X$ and $U$ a neighborhood of $x$ such that $f\rvert_U : U \rightarrow f(U)$ is a homeomorphism. There exists an open $O$ such that $x \in O \subseteq U$ so $f(x) \in f(O) \subseteq f(U)$ and $f(O) = f\rvert_U(O)$ is open because $f\rvert_U$ is a homeomorphism. Moreover, $f\rvert_O : O \rightarrow f(O)$ is a homeomorphism because $O \subseteq U$. So the open $O$ satisfies the first definition.

And the other implication is obvious so the two definitions would be equivalents.

Can someone explain me where is my mistake?

EDIT: So we should ask that for every $x \in X$ there is a neighborhood $U$ such that $f(U)$ is a neighborhood of $f(x)$ and $f\rvert_U:U→f(U)$ is a homeomorphism?

EDIT2: If we ask that for every $ \in $ there is a neighborhood $U$ such that $f(U)$ is a neighborhood of $f(x)$ and $f\rvert_U: U \rightarrow f(U)$ is a homeomorphism :

Let $O_X$ an open of $X$ such that $x \in O_X \subseteq U$ and $O_Y$ an open of $Y$ such that $f(x) \in O_Y \subseteq f(U)$ then we consider the open $O_X \cap f^{-1}(O_Y)$ whose image $f(O_X) \cap O_Y$ is an open of $Y$ (because it is an open of $O_Y$ and $O_Y$ is open) and we have a homeomorphism $\overline{f} : O_X \cap f^{-1}(O_Y) \rightarrow f(O_Y) \cap O_Y$.

However I don't understand the terminology of étale space because we say that $X$ is an étale space of $Y$ if there is a local homeomorphism $f : X \rightarrow Y$ so an open subset of $Y$ would be an étale space. Am I wrong? Thank you

C2H6
  • 41
  • 1
    I'll have to think about a counter example, but the flaw in your argument is that while $f(O)$ is open in $f(U)$ by definition of a homeomorphism, that doesn't guarantee that it is open in $X$, so there is a distinction in the definitions. – Keen Oct 21 '23 at 15:28
  • 1
    You get a different notion with your modified definition. A better name for your concept is a "local embedding." – Moishe Kohan Oct 21 '23 at 15:31
  • 1
    The updated definition is indeed equivalent to the standard one. What did you try to prove this equivalence? – Moishe Kohan Oct 21 '23 at 21:21

2 Answers2

4

The open $O$ most certainly does not satisfy the first condition (in general). $O$ is open in $f(U)$; it is not necessarily open in $Y$!

For a really simple example, we know the obvious inclusion $\Bbb R\hookrightarrow\Bbb R^2$ is an embedding. It is a (local) homeomorphism with its image but it is extremely false that this is a local homeomorphism; codomains matter. Since $\Bbb R\times\{0\}$ is not open in $\Bbb R^2$, there is no reason to expect the local homeomorphism condition to hold.

FShrike
  • 40,125
1

The flaw in your argument is that while f(O) is open in f(U) by definition of a homeomorphism, that doesn't guarantee that it is open in X, so there is a distinction in the definitions.

Consider for example the map $f$ from the closed interval $[0,1]$ to the open interval $(-2,2)$ that to $x$ associates $x$ itself. The image of any open neighborhood of $1$ by $f$ cannot contain any interval of the form $(1-\epsilon , 1+\epsilon )$ and thus cannot be open in $(-2,2)$. Yet the map $f$ is clearly a homeomorphism onto its image, since it is simply the identity map.

Keen
  • 1,125