Proposition 4.7.3. Let $X$ be a topological space, $\seqi{Y}$ be topological spaces, $\seqi{f}$ where $f_{i} \in C(X; Y_{i})$ for each $i \in I$, then
There exists a unique $f \in C(X; \prod_{i \in I}Y_{i})$ such that the following diagram commutes
\[\xymatrix{ & \prod_{i \in I} Y_i \ar@{->}[rd]^{\pi_i} & \\ X \ar@{->}[rr]_{f_i} \ar@{->}[ru]^{f} & & Y_i }\]for all $i \in I$.
If $X$ is T0 and equipped with the initial topology induced by $\seqi{f}$, then $f$ is an embedding.
Proof. (1): By (U) of Definition 4.7.1.
(2): Let $x, y \in X$ with $x \ne y$. Assume without loss of generality that there exists $U \in \cn(x)$ such that $y \not\in U$. Let $J \subset I$ finite, $\seqj{U}$, and $\seqj{V}$ such that
then $f(x) \in \bigcap_{j \in J}\pi_{j}^{-1}(U_{j})$ but $f(y) \not\in \bigcap_{j \in J}\pi_{j}^{-1}(U_{j})$. Thus $f$ is injective.
Let $U \subset X$ be open. Assume without loss of generality that there exists $J \subset I$ finite and $\bracs{U_j}_{j \in J}$ such that $U = \bigcap_{j \in J}f_{j}^{-1}(U_{j})$. In which case, $\bigcap_{j \in J}\pi_{j}^{-1}(U_{j})$ is open in $\prod_{i \in I}Y_{i}$ and
is a relatively open set.$\square$