# Contents

## Idea

An embedding of topological spaces is a continuous function which is a homeomorphism onto its image.

## Definition

###### Definition

(embedding of topological spaces)

Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be topological spaces. A continuous function $f \;\colon\; X \longrightarrow Y$ is called an embedding if
in its image factorization

$f \;\colon\; X \overset{\phantom{A}\simeq\phantom{A}}{\longrightarrow} f(X) \overset{\phantom{AAA}}{\hookrightarrow} Y$

with the image $f(X) \hookrightarrow Y$ equipped with the subspace topology, then $X \to f(X)$ is a homeomorphism.

## Properties

###### Proposition

(open/closed continuous injections are embeddings)

A continuous function $f \colon (X, \tau_X) \to (Y,\tau_Y)$ which is

1. an open map or a closed map

is an embedding (def. 1).

This is called a closed embedding if the image $f(X) \subset Y$ is a closed subset.

###### Proof

If $f$ is injective, then the map onto its image $X \to f(X) \subset Y$ is a bijection. Moreover, it is still continuous with respect to the subspace topology on $f(X)$. Now a bijective continuous function is a homeomorphism precisely if it is an open map or a closed map (by this prop.). But the image projection of $f$ has this property, respectively, if $f$ does (by this prop.).

###### Proposition

Embeddings of topological spaces are precisely the regular monomorphisms in the category Top of all topological spaces.

For proof see at Top this proposition.

###### Lemma

In $Top$, the pushout $j$ of a (closed/open) embedding $i$ along any continuous map $f$,

$\array{ A & \stackrel{i}{\hookrightarrow} & B \\ \mathllap{f} \downarrow & po & \downarrow \mathrlap{g} \\ C & \underset{j}{\hookrightarrow} & D, }$

is again a (closed/open) embedding.

For proof see at subspace topology here.

###### Proposition

(injective proper maps to locally compact spaces are equivalently the closed embeddings)

Let

1. $X$ be a topological space

2. $Y$ a locally compact topological space

3. $f \colon X \to Y$ be a continuous function.

Then the following are equivalent

1. $f$ is an injective proper map,

2. $f$ is a closed embedding (def. 1).

###### Proof

In one direction, if $f$ is an injective proper map, then since proper maps to locally compact spaces are closed, it follows that $f$ is also closed map. The claim then follows since closed injections are embeddings, and since the image of a closed map is closed.

Conversely, if $f$ is a closed embedding, we only need to show that the embedding map is proper. So for $C \subset Y$ a compact subspace, we need to show that the pre-image $f^{-1}(C) \subset X$ is also compact. But since $f$ is an injection (being an embedding), that pre-image is just the intersection $f^{-1}(C) \simeq C \cap f(X)$.

To see that this is compact, let $\{V_i \subset X\}_{i \in I}$ be an open cover of the subspace $C \cap f(X)$, hence, by the nature of the subspace topology, let $\{U_i \subset Y\}_{i \in I}$ be a set of open subsets of $Y$, which cover $C \cap f(X) \subset Y$ and with $V_i$ the restriction of $U_i$ to $C \cap f(X)$. Now since $f(X) \subset Y$ is closed by assumption, it follows that the complement $Y \setminus f(X)$ is open and hence that

$\{ U_i \subset Y \}_{i \in I} \sqcup \{ Y \setminus f(X) \}$

is an open cover of $C \subset Y$. By compactness of $C$ this has a finite subcover. Since restricting that finite subcover back to $C \cap f(X)$ makes the potential element $Y \setminus f(X)$ disappear, this restriction is a finite subcover of $\{V_i \subset C \cap f(X)\}$. This shows that $C \cap f(X)$ is compact.

Revised on June 28, 2017 15:31:38 by Urs Schreiber (46.183.103.17)