Product topology

The product topology is the canonical way of defining a topology on the Cartesian product of spaces. Let ${(𝑋 𝛼, T 𝛼)} 𝛼 \in 𝐴$ be an arbitrary collection of topological spaces with cartesian product

𝑋 = \prod 𝛼 \in 𝐴 𝑋 𝛼

and $𝜋 𝛼 : 𝑋 ↠ 𝑋 𝛼$ as projections. The product topology on $𝑋$ is the coarsest topology on $𝑋$ for which all projections $𝜋 𝛼$ are continuous.¹ #m/def/topology Thus it has Topological subbasis

A 𝑋 = {𝜋 - 1 𝛼 𝑈 : 𝑈 \in T 𝛼 : 𝛼 \in 𝐴}

Further characterisations

Explicit

The explicit characterisation is a little clunky due to the quirks of uncountable cartesian products. The product topology may be defined with the following topological basis #m/thm/topology

B 𝑋 = {\prod 𝛼 \in 𝐴 𝑈 𝛼 : 𝑈 𝛼 \in T 𝛼 w h e r e 𝑈 𝛼 \neq 𝑋 𝛼 f o r f i n i t e l y m a n y 𝛼}

Proof of basis

It follows from the first characterisation that the following forms a Topological subbasis

A 𝑋 = {𝜋 - 1 𝛼 𝑈 : 𝑈 \in T 𝛼 : 𝛼 \in 𝐴} = {\prod 𝛽 \in 𝐴 {𝑈 𝛽 = 𝛼 𝑋 𝛼 𝛽 \neq 𝛼 : 𝑈 \in T 𝛼 : 𝛼 \in 𝐴}

When this is completed to a Topological basis via finite intersections, one obtains the explicit characterisation above.

Universal property for the product topoloogy

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

Proof

We will first prove that the product topology satisfies the universal property. Let ${𝑋 𝛼, T 𝛼} 𝛼 \in 𝐴$ be topological spaces and let $𝑋 = \prod 𝛼 \in 𝐴 𝑋 𝛼$ be the cartesian product endowed with the product topology $T 𝑋$ . Let $(𝑍, T 𝑍)$ be a topological space, and $𝑓 : 𝑍 \to 𝑋$ be a function. If $𝑓$ is continuous, then so are the compositions $𝜋 𝛼 𝑓$ of continuous functions for all $𝛼 \in 𝐴$ . Now suppose $𝜋 𝛼 𝑓 : 𝑍 \to 𝑋 𝛼$ is continuous for all $𝛼 \in 𝐴$ . We use the method of Proving continuity with a subbasis. Let $𝑈 \in A 𝑋$ . Then $𝑈 = 𝜋 - 1 𝛼 𝑉$ for some $𝛼 \in 𝐴$ and $𝑉 \in 𝑋 𝛼$ . Since $𝜋 𝛼 𝑓$ is continuous, $𝑓 - 1 𝑈 = (𝜋 𝛼 𝑓) - 1 𝑉 \in T 𝑍$ . Thus the preïmage $𝑓 - 1 𝑈$ of every subbasic open set $𝑈 \in A 𝑋$ is open, whence $𝑓$ is continuous. Therefore $𝑓$ is continuous iff $𝜋 𝛼 𝑓$ is 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 $𝑓 = i d 𝑋 : (𝑋, T 𝑋) \to (𝑋, T')$ . Then since $𝜋 𝛼 i d 𝑋 = 𝜋 𝛼 : (𝑋, T 𝑋) \to 𝑋 𝛼$ is continuous for all $𝛼 \in 𝐴$ , so is $i d 𝑋 : (𝑋, T 𝑋) \to (𝑋, T')$ , wherefore $T'$ is coarser than $T 𝑋$ . Now let $(𝑍, T 𝑍) = (𝑋, T')$ and $𝑓 = i d 𝑋 : (𝑋, T') \to (𝑋, T')$ . Since $i d 𝑋$ is continuous, so too is $𝜋 𝛼 𝑓$ for all $𝛼 \in 𝐴$ . But $T 𝑋$ is the coarsest topology on $𝑋$ such that $𝜋 𝛼 𝑓$ is continuous for all $𝛼 \in 𝐴$ , so $T 𝑋 = T'$ .

Spaces constructed as products

Real coördinate space as products of $ℝ$ with the standard topology, e.g. $ℝ 2 = ℝ \times ℝ$ .
Torus topology $𝕋 1 = 𝕊 1 \times 𝕊 1$

Properties

Continuous maps from the product topology are continuous in each argument
Canonical projections are open

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

2020, Topology: A categorical approach, pp. 30–31 ↩