This question comes from a beginner undergraduate who just read the definitions and likes to wonder things, so it is possible I'm asking too much. I appreciate your patience.
The setup
Let $(X_{i})_{i\in I}$ be a family of topological spaces and $(f_{i}\colon X\to X_{i})_{i\in I}$ a family of maps defined on $X$.
We put the initial (or weak) topology on $X$. We hold this for a while.
Now we take the cartesian product $\prod X_{i}$ and consider the product topology on it.
From the universal property of the product topology, there is a unique continuous map \begin{equation} X\to \prod X_{i} \end{equation} making the 'usual' diagrams commute.
Wikipedia says here that this map is known as the 'evaluation map'. They also say that this map is injective if and only if the family $(f_{i})_{i\in I}$ separates points.
The question
Do we know conditions under which this will be surjective? What about a homeomorphism?
If this is too general, do we know additional conditions which give such a criterion?
I doubt there is any useful condition for the map $f:X\to\prod X_i$ to be surjective that isn't just a paraphrase of the definition. In practice, it very rarely is surjective.
However, if $f$ is a bijection, then it automatically is a homeomorphism: this follows almost immediately from the definition of the product topology and the initial topology. Indeed, the initial topology is defined to be generated by sets of the form $f_j^{-1}(U)=f^{-1}(\pi_j^{-1}(U))$ where $\pi_j:\prod X_i\to X_j$ is the $j$th projection and $U\subseteq X_j$ is open. The product topology on $\prod X_i$ is defined to be generated by sets of the form $\pi_j^{-1}(U)$. So when $f$ is a bijection, the sets generating the topology on $X$ are "the same" as the sets generating the topology on $\prod X_i$ (just pulled back by the bijection $f$), so $f$ is a homeomorphism. More generally, a similar argument shows that $f$ is automatically a homeomorphism onto its image if it is injective.