It's well known that if $G$ is an Abelian group and $A, B$ are subgroups such that $A \cap B = \{0\}$, then there is an isomorphism between the external direct sum $A \bigoplus B$ and internal direct sum $A\oplus B$ given by $$\nabla(a, b) = a+b$$
My question is: If $G$ is a topological Abelian group, then is $\nabla$ also a homeomorphism if we consider the subspace topology on the internal direct sum and the product topology on the external one? It's easy to see that it is continuous, the problem lies in its inverse. For the particular case I'm interested in, $G$ is also compact, but I'm not sure whether that makes any difference.
In the product topology on $A \bigoplus B$, the subspace $A$ is always closed as long as $B$ is Hausdorff.
Consider $G:=\prod_{n \in \Bbb Z} \Bbb Z/2\Bbb Z$ with the product topology. Then $A=\bigoplus_{n \in \Bbb Z} \Bbb Z/2\Bbb Z \subset G$ has a complement $B$ by linear algebra (everything is a vector space over the field with two elements here). But the canonical bijective continuous homomorphism $A \oplus B \to G$ is not a homeomorphism, because $A$ is not closed in $G$, but it is in $A \oplus B$.
By Tychonoff, $G$ is compact.