nLab embedding of topological spaces

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, we have that $X \to f(X)$ 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 injective function

2. an open map or a closed map

is an embedding (def. ).

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. ).

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.

Related concepts

• embedding

• embedding of smooth manifolds