For consider a family of topological abelian groups $\{G_\alpha\}_{\alpha \in J}$ for some index set $J$. I sought for an easy description of the topology that the direct sum $G = \bigoplus_{\alpha \in J} G_\alpha$ carries. Here's my attempt.
Note that each coset $xG_\alpha \subset G$ can be topologized by copying the topology of $G_\alpha$. This is well-defined over choices of $x$ because the topology carried by $G_\alpha$ is translation-invariant.
Let $\mathcal{T}$ the final topology over $G$ induced by all inclusion maps $i_{\alpha, x} : xG_\alpha \to G$. That is, a subset $U \subset G$ shall be open if and only if its inverse image $i_{\alpha, x}^{-1}[U]$ is open in $xG_\alpha$ for each $\alpha \in J$ and $x \in G$.
Let me prove that, under this topology, the group operations (binary operation and inversion) are continuous. It suffices to prove that $f : G \times G \to G$ defined by $f(x,y) = x^{-1} y$ is continuous. Since $\mathcal{T}$ is translation-invariant, it also suffices to prove that $f$ is continuous at $(e,e)$. Let $V \subset G$ be an open neighborhood of $e \in G$. Let $U_\alpha$ denote $i_{\alpha,e}^{-1}[V]$. Then $U_\alpha \subset G_\alpha$ is an open neighborhood of $e \in G_\alpha$. Since $G_\alpha$ is a topological group, it's possible to pick a neighborhood $U_\alpha'$ of $e \in G_\alpha$ such that, for every $x,y \in U_\alpha'$, $x^{-1} y \in U_\alpha$ holds.
Now consider the set $U = \bigcup_{1, 2, \cdots, n \in J} \{i_{1, e}(x_1) i_{2, e}(x_2) \cdots i_{n, e}(x_n) : x_1 \in U'_1, x_2 \in U'_2, \cdots, x_n \in U'_n\}$. Then $U$ is an open neighborhood of $e \in G$ such that $f[U \times U] \subset V$, and the proof is complete.
It should be easy to check the universal property, but how can I modify this proof for general (might-be non-abelian) topological groups and free products? I guess, instead of $xG_\alpha$, I'd need $xG_\alpha y$, and in the last step, the indices $1, 2, \cdots, n \in J$ would need not be pairwise different.
