When learning about the concept of complemented subspaces of a Banach space, I'm curious with the following question:
Let $X$ be a vector space over $\mathbb{C}$ or $\mathbb{R}$ and let $E$ and $F$ be algebraic complemented subspaces with trivial intersection, such that $X=\ E\oplus F$. If defining a norm $\|\cdot\|$ on $X$, and equipping $E$ and $F$ with their norm-induced subspace topologies, respectively, is it always true that, as topological spaces, $X\simeq E\times F$?
A closely related proposition in terms of linear projections can be stated as follows.
Let $(X, \|\cdot\|)$ be a normed space over $\mathbb{C}$ or $\mathbb{R}$, and $P: X\to X$ a linear projection such that $P^2=P$. Then, do we always have a topological direct sum result $X\simeq \text{ran}(P) \oplus \text{Ker}(P)$, where the range and kernel are equipped with their subspace topologies, respectively?
Here I'm not assuming continuity of $P$ or whether the range or kernel are closed subspaces.
Background:
I'm trying to understand, on Banach spaces, why there can be discontinuous projections with a closed range.
In a Banach space $X$, if a subspace $E\subset X$ is closed, then its quotient space $X/E$ is a Banach space (complete) under the quotient norm. Purely algebraically, we would then have $X \simeq E \oplus (X/E)$. But, for a subspace $F\subset X$ algebraically complemented to $E$, we would also have algebraic direct sum result $X=E\oplus F$. This seems to suggest, the natural algebraic isomorphism from $F$ to $(X/E)$ should also be a homeomorphism. But, since not every closed subspace $E$ is (alg + top) complemented, that shouldn't be true.
My question is to help me understand this confusion. If the propositions in my question are true, then I can understand, even when $\text{ran}(P)$ is closed, the algebraic isomorphism from $X/\text{ran}(P)$ to $\text{Ker}(P)$ can fail to be a homeomorphism. So the kernel can still be non-closed.
I already know the following two facts/propositions about linear projections on Banach spaces.
If a projection $P$ is continuous, then $\text{ran}(P)$ is closed, since it is the kernel of $I-P$.
If a projeciton $P$ has closed range and closed kernel, it is continuous. This can be proved since everything is Banach (complete) and so we can freely apply the Open Mapping Thm/Inverse Mapping Thm wherever possible. See here.
Thanks in advance.
Update:
The conclusive answer should be the following.
Let $X$ be a normed vector space and $E\subset X$ a closed subspace, with its purely algebraic complemented subspace $F\subset X$ such that $X=E\oplus F$. Then, the original norm topology $\tau_X$ of the whole space, the product topology of the subspaces $\tau|_E \times \tau|_F$, and the product topology with the quotient space $\tau|_E \times \tau_{X/E}$, are in general all different.
Best Answer
If $X=E\times F$, let $p:X\to E$ be the projection $p:(x,y)\longmapsto x$. Let $V\subset E$ be open. Then $$ p^{-1}(V)=V\times F. $$ Because the topology is the product topology, $V\times F$ is open. Hence $p$ is continuous.
This means that if your projection $P$ is not continuous, the direct sum $\operatorname{ran}P\oplus\ker P$ is not topological.
It is easy to produce unbounded projections. Indeed, given any infinite-dimensional Banach space $X$, let $g$ be an unbounded linear functional. Fix nonzero $z\in X$ and define $$ Px=g(x)z. $$ Then $P$ is an unbounded rank-one projection, and $\ker P=\ker g$ is dense.