Proof of the universal property of the product topology

Proof

We prove the two directions separately. Let $\tau$ be the product topology on $X$, and let $\mathcal{S} = \set{\pi_i^{-1}(U) : i \in I, \ U \text{ open in } X_i}$.

1. By definition, $\tau$ is the topology generated by the subbasis $\mathcal{S}$. In particular $\mathcal{S} \subseteq \tau$.
2. Each projection $\pi_i$ is continuous. Indeed, for $U$ open in $X_i$, the preimage $\pi_i^{-1}(U)$ is in $\mathcal{S} \subseteq \tau$.
3. Let $f$ be continuous. Then each $\pi_i \circ f$ is a composition of continuous maps, so it is continuous.
4. Conversely, let each $\pi_i \circ f$ be continuous. Let $\sigma$ be the set of all $W \subseteq X$ such that $f^{-1}(W)$ is open in $Z$.
5. We check that $\sigma$ is a topology on $X$. First, $f^{-1}(\varnothing) = \varnothing$ and $f^{-1}(X) = Z$ are open in $Z$. So $\varnothing, X \in \sigma$.
6. For a family $(W_k)$ in $\sigma$, $f^{-1}(\bigcup_k W_k) = \bigcup_k f^{-1}(W_k)$, a union of open sets of $Z$. So $\bigcup_k W_k \in \sigma$.
7. For $W, W' \in \sigma$, $f^{-1}(W \cap W') = f^{-1}(W) \cap f^{-1}(W')$, which is open in $Z$. So $W \cap W' \in \sigma$.
8. Let $i \in I$ and $U$ open in $X_i$. Then $f^{-1}(\pi_i^{-1}(U)) = (\pi_i \circ f)^{-1}(U)$, by the definition of the composition. This set is open in $Z$, so $\pi_i^{-1}(U) \in \sigma$.
9. So $\sigma$ is a topology on $X$ that contains $\mathcal{S}$. Since $\tau$ is the smallest such topology, $\tau \subseteq \sigma$.
10. Thus $f^{-1}(W)$ is open in $Z$ for each $W \in \tau$, and $f$ is continuous.