Because the direct sum of two topological vector spaces (TVS) equipped with the product topology is automatically a TVS, I wonder whether the converse is true: the topology of a TVS is the same as the product topology of a pair of its complementary vector subspaces (equipped with the subspace topology).
I found that this statement is equivalent to this: in a TVS, the cylinder generated by a cross section of an open set is an open set. Formally, let $V$ be a TVS and $X,Y$ be a pair of its complementary vector subspaces. Then, $$\forall A\in\tau_V:\left(A\cap X\right)+Y\in\tau_V.$$
Intuitively this is true in $\mathbb R^n$. Can this be proved in any TVS?
Another way of asking the same question is "is every linear projection continuous?" That's not true. Here's a counter example.
Suppose we have a Hilbert space $H$. A one-dimensional subspace $V$ is spanned by the vector $v=(1,1/2,1/3,\ldots)$. To choose the complementary subspace, we pick a basis: first, make sure that the basis contains a sequence $v_1,v_2,\ldots$ converging to $v$, and then extend (by AC) arbitrarily to some set $B$ such that $B$ together with $v$ forms a basis for $H$. Then, the projection $H\to V$ which maps $B$ to $0$ is not continuous because it maps all $v_i$ to $0$ but maps $v$ to $v\ne0$.