Contents

# Contents

## Idea

A continuous map $f : X \to Y$ between topological spaces is called a local homeomorphism if restricted to a neighbourhood of every point in its domain it becomes a homeomorphism onto its image which is required to be open.

One also says that this exhibits $X$ as an étale space over $Y$.

Notice that, despite the similarity of terms, local homeomorphisms are, in general, not local isomorphisms in any natural way. See the examples below.

## Definition

###### Definition

A local homeomorphism is a continuous map $p \colon E \to B$ between topological spaces (a morphism in Top) such that

• for every element $e \in E$, there is an open subset $U \ni e$ whose image $p_*(U)$ is open in $B$ and the restriction of $p$ to $U$ is a homeomorphism $p|_U \colon U \to p_*(U)$,

or equivalently

• for every $e \in E$, there is a neighbourhood $U$ of $e$ such that the image $p_*(U)$ is a neighbourhood of $p(e)$ and $p|_U: U \to p_*(U)$ is a homeomorphism.

## Examples

For $X$ any topological space and for $S$ any set regarded as a discrete space, the projection

$X \times S \to X$

is a local homeomorphism.

For $\{U_i \to Y\}$ an open cover, let

$X := \coprod_i U_i$

be the disjoint union space of all the pathches. Equipped with the canonical projection

$\coprod_i U_i \to Y$

this is a local homeomorphism.

In general, for every sheaf $A$ of sets on $Y$; there is a local homeomorphism $X \to Y$ such that over any open $U \hookrightarrow X$ the set $A(U)$ is naturally identified with the set of sections of $Y \to X$. See étale space for more on this.

## Properties

###### Proposition

A local homeomorphism is an open map.

###### Proof

Let $f \colon X \to Y$ be a local homeomorphism and $U \subset X$ an open subset. We need to see that the image $f(U) \subset Y$ is an open subset of $Y$. For this we may equivalently show that each $y \in f(U)$ has an open neighbourhood inside $f(U)$.

But since any function is surjective onto its image, there exists $x \in U$ with $f(x) = y$. By local homeomorphy of $f$, this $x \in X$ has an open neighbourhood $U_x \subset X$ with $f_{|U_x} \colon U_x \to f(U_x)$ a homeomorphism. Since $U \cap U_x \subset U_x$ is an open neighbourhood of $x$ in $U_x$, the homeomorphy of $f_{|U_x}$ implies that $f(U \cap U_x) \subset f(U)$ is an open neighbourhood of $f(x) = y$.

Last revised on June 30, 2022 at 02:05:19. See the history of this page for a list of all contributions to it.