TVS = topological vector space. Any subspace of a TVS is a TVS in the induced-topology sense.
DEFINITION
For TVS spaces ${\mathbb{V}}_1\subset\mathbb V$, a TVS subspace ${\mathbb{V}}_2\subset\mathbb V$ is a topological complement of ${\mathbb{V}}_1$ in $\mathbb V$, if $\mathbb V$ is their direct sum both algebraically and topologically. This implies that $$ {\mathbb{V}}_1\oplus {\mathbb{V}}_2\,=\,{\mathbb{V}}~\,, $$ and the following addition map is a homeomorphism: $$ {\mathbb{V}}_1\times{\mathbb{V}}_2\,\longrightarrow\,{\mathbb{V}}~~,\qquad \left(\, v\in{{\mathbb V}}_1\,,~\,v^{\,\prime}\in{{\mathbb V}}_2\,\right)\,\longmapsto~ (v+v^{\,\prime})\in\mathbb V~~.\\~\\ $$
While the extension of the above construction to a finite direct sum of TVS spaces is straightforward, splitting $\mathbb V$ into an infinite direct sum of TVS subspaces is not that trivial, especially when the number of summands is uncountable.
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QUESTION
For the splitting $$ {\mathbb V}=\bigoplus_{i\in\cal I}{\mathbb V}_i $$ to be not only algebraic but also topological, would it be sufficient to impose the condition that the map $$ \{{\mathbb{V}}_i\}_{i\in{\cal I}}\longrightarrow\mathbb{V} $$ exists and is a homeomorphism?
Would this work (a) for a countable sum, and (b) for an uncountable sum (direct integral) of subspaces?