Coproduct topology

The coproduct topology or sum topology is the canonical way of defining a topology on the Disjoint union of spaces. Let ${(𝑋 𝛼, T 𝛼)} 𝛼 \in 𝐴$ be an arbitrary collection of topological spaces with disjoint union

𝑋 = ∐ 𝛼 \in 𝐴 𝑋 𝛼

and $𝜄 𝛼 : 𝑋 𝛼 ↣ 𝑋$ as canonical inclusions. The coproduct topology on $𝑋$ is the finest topology on $𝑋$ for which all inclusions $𝜄 𝛼$ are continuous.¹ #m/def/topology

T 𝑋 = {𝑈 \subseteq 𝑋 : 𝜄 - 1 𝛼 𝑈 \in T 𝛼 \forall 𝛼 \in 𝐴}

Further characterisations

Explicit

The open sets in the coproduct topology correspond exactly to the unions of images of open sets in the constituent topologies, i.e.

T 𝑋 = {⋃ 𝛼 \in 𝐴 𝜄 𝛼 𝑈 𝛼 : 𝑈 𝛼 \in T 𝛼} = {∐ 𝛼 \in 𝐴 𝑈 𝛼 : 𝑈 𝛼 \in T 𝛼}

Bases for the coproduct topology

The images of open sets form a topological basis:

B 𝑋 = {𝜄 𝛼 𝑈 𝛼 : 𝑈 𝛼 \in T 𝛼 : 𝛼 \in 𝐴}

It follows that if $B 𝛼$ is a basis of $𝑋 𝛼$ for each $𝛼 \in 𝐴$ , the following is also a basis of $𝑋$ :

B' 𝑋 = {𝜄 𝛼 𝑈 𝛼 : 𝑈 𝛼 \in B 𝛼 : 𝛼 \in 𝐴}

Universal property for the coproduct topology

For every topological space $(𝑍, T 𝑍)$ and function $𝑓 : 𝑋 \to 𝑍$ , then $𝑓$ is continuous iff $𝑓 𝜄 𝛼 : 𝑋 𝛼 \to 𝑍$ is continuous for all $𝛼 \in 𝐴$ . #m/thm/topology

Proof

First we will prove that the coproduct topology as characterised above satisfies the universal property. Let ${(𝑋 𝛼, T 𝛼)} 𝛼 \in 𝐴$ be topological spaces and let $𝑋 = ∐ 𝛼 \in 𝐴$ be their disjoint union endowed with the coproduct topology and with canonical inclusions $𝜄 𝛼 : 𝑋 𝛼 ↣ 𝑋$ . Let $(𝑍, T 𝑍)$ be some topological space, and let $𝑓 : 𝑋 \to 𝑍$ be a function. If $𝑓$ is continuous, then so are the compositions $𝑓 𝜄 𝛼$ of continuous functions for all $𝛼 \in 𝐴$ . Now suppose $𝑓 𝜄 𝛼 : 𝑋 𝛼 \to 𝑍$ is continuous for all $𝛼 \in 𝐴$ , and let $𝑈 \in T 𝑍$ . Then $(𝑓 𝜄 𝛼) - 1 𝑈 = 𝜄 - 1 𝛼 𝑓 - 1 𝑈 \in T 𝛼$ for all $𝛼 \in 𝐴$ , whence $𝑓 - 1 𝑈 \in T 𝑋$ . Thus $𝑓$ is continuous. Therefore $𝑓$ is continuous iff $𝑓 𝜄 𝛼$ are continuous for all $𝛼 \in 𝐴$ .

Now let $T'$ be a topology on $𝑋$ satisfying the same universal property. In particular, let $(𝑍, T 𝑍) = (𝑋, T 𝑋)$ and $𝑓 = i d 𝑋 : (𝑋, T') \to (𝑋, T 𝑋)$ . Then since $i d 𝑋 𝜄 𝛼$ is continuous for all $𝛼 \in 𝐴$ , so is $i d 𝑋 : (𝑋, T') \to (𝑋, T 𝑋)$ , wherefore $T'$ is finer than $T 𝑋$ Now let $(𝑍, T 𝑍) = (𝑋, T 𝑋)$ and $𝑓 = i d 𝑋 : (𝑋, T') \to (𝑋, T')$ . Since $i d 𝑋$ is continuous, so too is $i d 𝑋 𝜄 𝛼 = 𝜄 𝛼$ for all $𝛼 \in 𝐴$ . But $T 𝑋$ is the finest topology for which all $𝜄 𝛼$ are continuous, so $T' = T 𝑋$

Spaces constructed as coproducts

The Discrete topology is the coproduct of all its points viewed as singletons.

#state/tidy | #lang/en | #SemBr

2020, Topology: A categorical approach, pp. 32–33 ↩