Universal Property of the Product Topology

The product topology is uniquely characterized by a universal property: it is the coarsest topology making all projections continuous. This characterization explains why the product topology (rather than the box topology) is the "correct" topology on a Cartesian product.

Statement

Theorem3.6Universal Property of the Product Topology

Let $\{(X_\alpha, \tau_\alpha)\}_{\alpha \in A}$ be a family of topological spaces and let $P = \prod_{\alpha \in A} X_\alpha$ carry the product topology. For each $\alpha$ , let $\pi_\alpha: P \to X_\alpha$ be the projection map.

Then the product topology is the unique topology on $P$ satisfying the following universal property: for any topological space $Z$ and any family of continuous maps $\{f_\alpha: Z \to X_\alpha\}_{\alpha \in A}$ , there exists a unique continuous map $f: Z \to P$ such that $\pi_\alpha \circ f = f_\alpha$ for all $\alpha$ .

$\begin{array}{ccc} & & X_\alpha \\ & \nearrow^{f_\alpha} & \uparrow^{\pi_\alpha} \\ Z & \xrightarrow{f} & P \end{array}$

The map $f$ is defined by $f(z) = (f_\alpha(z))_{\alpha \in A}$ .

Proof

Existence of $f$ : Define $f: Z \to P$ by $f(z)(\alpha) = f_\alpha(z)$ for each $\alpha \in A$ . Then $\pi_\alpha(f(z)) = f(z)(\alpha) = f_\alpha(z)$ , so $\pi_\alpha \circ f = f_\alpha$ .

Uniqueness of $f$ : If $g: Z \to P$ satisfies $\pi_\alpha \circ g = f_\alpha$ for all $\alpha$ , then for every $z \in Z$ and every $\alpha$ , $\pi_\alpha(g(z)) = f_\alpha(z) = \pi_\alpha(f(z))$ . Since the projections separate points, $g(z) = f(z)$ for all $z$ .

Continuity of $f$ : We use the subbasis criterion. The subbasis for the product topology consists of sets $\pi_\alpha^{-1}(U_\alpha)$ where $U_\alpha$ is open in $X_\alpha$ . We have: $f^{-1}(\pi_\alpha^{-1}(U_\alpha)) = (\pi_\alpha \circ f)^{-1}(U_\alpha) = f_\alpha^{-1}(U_\alpha),$ which is open in $Z$ since $f_\alpha$ is continuous.

Characterization of the topology: We show the product topology is the only topology on $P$ satisfying the universal property.

Let $\tau$ be any topology on $P$ with the universal property. Taking $Z = (P, \tau_{\text{prod}})$ with $f_\alpha = \pi_\alpha$ (continuous in the product topology), the universal property for $\tau$ gives that $\operatorname{id}: (P, \tau_{\text{prod}}) \to (P, \tau)$ is continuous, so $\tau \subseteq \tau_{\text{prod}}$ .

Conversely, taking $Z = (P, \tau)$ with $f_\alpha = \pi_\alpha$ (continuous in $\tau$ , since the universal property requires the projections to be continuous), the universal property for $\tau_{\text{prod}}$ gives that $\operatorname{id}: (P, \tau) \to (P, \tau_{\text{prod}})$ is continuous, so $\tau_{\text{prod}} \subseteq \tau$ .

Therefore $\tau = \tau_{\text{prod}}$ .

■

Categorical Perspective

RemarkProduct in the Category $\mathbf{Top}$

The universal property states precisely that $(P, \{\pi_\alpha\})$ is a product in the category $\mathbf{Top}$ of topological spaces and continuous maps. In categorical language: $\operatorname{Hom}_{\mathbf{Top}}(Z, \textstyle\prod_\alpha X_\alpha) \cong \prod_\alpha \operatorname{Hom}_{\mathbf{Top}}(Z, X_\alpha)$ naturally in $Z$ . The isomorphism sends $f$ to $(\pi_\alpha \circ f)_\alpha$ .

This is a general construction in category theory. The product topology is the unique topology that makes the Cartesian product into a categorical product.

Applications

ExampleContinuous Maps into $\mathbb{R}^n$

A function $f: X \to \mathbb{R}^n$ is continuous if and only if each component function $f_i = \pi_i \circ f: X \to \mathbb{R}$ is continuous. This is a direct consequence of the universal property.

For instance, $f: \mathbb{R} \to \mathbb{R}^3$ defined by $f(t) = (\cos t, \sin t, t)$ is continuous because $\cos$ , $\sin$ , and $\operatorname{id}$ are all continuous.

ExampleProduct Topology as Initial Topology

More generally, the product topology is a special case of the initial topology (or weak topology): given a family of maps $f_\alpha: X \to Y_\alpha$ , the initial topology on $X$ is the coarsest topology making all $f_\alpha$ continuous. Its subbasis is $\{f_\alpha^{-1}(U_\alpha) : \alpha \in A, U_\alpha \text{ open in } Y_\alpha\}$ .

The product topology on $\prod_\alpha X_\alpha$ is the initial topology for the projection maps $\pi_\alpha$ . The subspace topology on $A \subseteq X$ is the initial topology for the inclusion $\iota: A \hookrightarrow X$ .

Dually, the quotient topology is a special case of the final topology (or strong topology).

Theorem3.7Products Preserve Hausdorff

The product $\prod_\alpha X_\alpha$ is Hausdorff if and only if each $X_\alpha$ is Hausdorff.

Proof

( $\Rightarrow$ ): Each $X_\alpha$ is homeomorphic to a subspace of $\prod_\alpha X_\alpha$ (fix all other coordinates), and subspaces of Hausdorff spaces are Hausdorff.

( $\Leftarrow$ ): Let $\mathbf{x} \neq \mathbf{y}$ in $\prod_\alpha X_\alpha$ . Then $x_\beta \neq y_\beta$ for some $\beta$ . Since $X_\beta$ is Hausdorff, there exist disjoint open sets $U, V$ in $X_\beta$ with $x_\beta \in U$ , $y_\beta \in V$ . Then $\pi_\beta^{-1}(U)$ and $\pi_\beta^{-1}(V)$ are disjoint open sets in the product separating $\mathbf{x}$ and $\mathbf{y}$ .

■