nLab embedding of topological spaces

Redirected from "closed embedding of topological spaces".
Contents

Context

Topology

topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory

Introduction

Basic concepts

Universal constructions

Extra stuff, structure, properties

Examples

Basic statements

Theorems

Analysis Theorems

topological homotopy theory

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

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

with the image f(X)Yf(X) \hookrightarrow Y equipped with the subspace topology, we have that Xf(X)X \to f(X) is a homeomorphism.

Properties

Proposition

(open/closed continuous injections are embeddings)

A continuous function f:(X,τ X)(Y,τ Y)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)Yf(X) \subset Y is a closed subset.

Proof

If ff is injective, then the map onto its image Xf(X)YX \to f(X) \subset Y is a bijection. Moreover, it is still continuous with respect to the subspace topology on f(X)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 ff has this property, respectively, if ff 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 TopTop, the pushout jj of a (closed/open) embedding ii along any continuous map ff,

A i B f po g C j D,\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. XX be a topological space

  2. YY a locally compact topological space

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

Then the following are equivalent

  1. ff is an injective proper map,

  2. ff is a closed embedding (def. ).

Proof

In one direction, if ff is an injective proper map, then since proper maps to locally compact spaces are closed, it follows that ff 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 ff is a closed embedding, we only need to show that the embedding map is proper. So for CYC \subset Y a compact subspace, we need to show that the pre-image f 1(C)Xf^{-1}(C) \subset X is also compact. But since ff is an injection (being an embedding), that pre-image is just the intersection f 1(C)Cf(X)f^{-1}(C) \simeq C \cap f(X).

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

{U iY} iI{Yf(X)} \{ U_i \subset Y \}_{i \in I} \sqcup \{ Y \setminus f(X) \}

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

Last revised on February 2, 2021 at 12:13:58. See the history of this page for a list of all contributions to it.