Disjoint Union Space

Disjoint union of spaces $A\bigsqcup B$ is the coproduct in topology. See coproduct in Type theory: Example Coproduct type A+B

Definition

Let $(X_{\alpha }{)}_{\alpha \in A}$ be an indexed family of spaces, and let ${\coprod }_{\alpha \in A}{X}_{\alpha }$ denote their disjoint union. Then a subset $\mathcal{U}\subseteq {\coprod }_{\alpha \in A}{X}_{\alpha }$ is open in the disjoint union topology if $\mathcal{U}\cap X_{\alpha }$ is open for each $\alpha \in A$

$$\coprod _{\alpha \in A}{x}_{\alpha }:=\{(x,a)|x\in X_{\alpha }\}$$

Since we want ${\iota }_{\alpha }:X_{\alpha }\hookrightarrow {\coprod }_{\alpha \in A}{X}_{\alpha }$ to be continuous, for every open $\mathcal{U}\subseteq {\coprod }_{\alpha \in A}{X}_{\alpha }$ we must have ${\iota }_{\alpha }^{-1}(\mathcal{U})$ is open. ${\iota }_{\alpha }^{-1}(\mathcal{U})=\{x\in X_{\alpha }|(x,\alpha )\in \mathcal{U}\}=X_{\alpha }\times \{\alpha \}\cap \mathcal{U}=X_a\cap \mathcal{U}$ Note that $X_{\alpha }\times \{\alpha \}\cong X_{\alpha }$

In simpler words: a open subset of the disjoint union spaces is a union of open subset of each space.

Notation

$\hookrightarrow$ means inclusion map a map $\iota :A\hookrightarrow X$ indicates $A\subseteq X$ and elements are mapped to themselves

Characteristic Property

Let $Y$ be any space. A function $f:{\coprod }_{\alpha \in A}{X}_{\alpha }\to Y$ is continuous $⟺$ $f_{|X_{\alpha }}:X_{\alpha }\to Y$ is continuous for all $\alpha \in A$.

We try to prove the $⟸$ direction

Proof using universal property

By universal property of the category of topological spaces we know that there exists a unique continuous map $\mathcal{U}$ that makes this graph commute.

Title

Note that $\mathcal{U}$ is unique among continuous maps, NOT among all functions. Thus, we need to prove $f$ is $\mathcal{U}$

That is $\mathcal{U}\circ {\iota }_{\alpha }=f_{|X_{\alpha }}=f\circ {\iota }_{\alpha }$

$f$ satisfies the universal property for disjoint union viewed as a set, because $f$ makes the graph commute. This gives $\mathcal{U}=f$ as functions $⟹$ $f$ is also continuous

Note that this argument is almost the same as the arguments in Universal properties of Product space. Only the directions of the arrows are reversed

Proof using definitions

Let $U\subseteq Y$ open we need to prove $f^{-1}(U)$ is open Consider $f^{-1}(U)\cap X_{\alpha }$ $f^{-1}(U)\cap X_{\alpha }=\{x\in \coprod X_{\alpha }|f(x)\in U\wedge x\in X_{\alpha }\}$ $=\{x\in X_{\alpha }|f(x)\in U\}=f_{|X_{\alpha }}^{-1}(U)$ Since $f_{|X_{\alpha }}$ is continuous, $f_{|X_{\alpha }}^{-1}(U)$ is open $⟹f^{-1}(U)\cap X_{\alpha }$ is open for all $\alpha \in A$ $⟹f^{-1}(U)$ is open (by definition of coproduct)

Properties

• $C\subseteq {\coprod }_{\alpha \in A}{X}_{\alpha }$ is closed $⟺$ $C\cap X_{\alpha }$ is closed for all $\alpha \in A$
• Each injection ${\iota }_{\alpha }:X_{\alpha }\to \coprod X_{\alpha }$ is an embedding
  • a topological embedding means a map is a homeomorphism onto its image
  • $X_{\alpha }$ is homeomorphic to ${\iota }_{\alpha }(X_{\alpha })$
  • This is straightforward, because $x\mapsto (x,\alpha )$
• Taking disjoint unions preserves:
  • Hausdorff property
  • First countability
  • Second countability, provided $A$ is countable