Universal property of the product topology

Statement

Let $(X_i)_{i \in I}$ be a family of topological spaces, let $X = \prod_{i \in I} X_i$ have the product topology, and let $\pi_i \colon X \to X_i$ be the projections. Let $Z$ be a topological space and let $f \colon Z \to X$ be a function. Then $f$ is continuous if and only if the composition $\pi_i \circ f \colon Z \to X_i$ is continuous for each $i \in I$.

Remarks

Proof of the converse: preimages of the subbasic sets $\pi_i^{-1}(U)$ under $f$ are $(\pi_i \circ f)^{-1}(U)$, which are open. Preimages respect unions and intersections, so all preimages of open sets are open. Thus a map into a product is continuous if and only if all its coordinate functions are continuous. For example, $t \mapsto (\cos t, \sin t)$ is continuous from $\mathbb{R}$ to $\mathbb{R}^2$. The statement fails for the box topology.

Explanations

L1 human

To check that a map into a product is continuous, check each coordinate separately. This property characterises the product topology among all topologies on the product.