embedding of topological spaces in nLab

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:X⟶Yf \;\colon\; X \longrightarrow Y is called an embedding if
in its image factorization

f:X⟶A≃Af(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 X→f(X)X \to f(X) is a homeomorphism.

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.

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 C⊂YC \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)≃C∩f(X)f^{-1}(C) \simeq C \cap f(X).

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

{U i⊂Y} i∈I⊔{Y∖f(X)} \{ U_i \subset Y \}_{i \in I} \sqcup \{ Y \setminus f(X) \}

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