Decomposing the topology of a normed linear space as the product topology of its subspaces Let's take a normed linear space $V$ and its subspaces $U$ and $W$ such that $V = U\oplus W$. Now, $U$, $W$ are normed linear spaces in their own right. It's easily shown that $U\oplus W\cong U\times W$ as vector spaces via the map $(u, w)\mapsto u + w$ from $U\times W$ to $U\oplus W$.
Question: Is $f$ also a homeomorphism from $U\times V$ (under the product topology due induced by the topologies of $U$ and $V$) to $U\oplus W$ (which is just $V$)? I am able to show that $f$ is continuous, but fail to show its openness.

By the Cartesian product, I am denoting the external direct product of spaces.
Also, I did browse through this question. But it basically asks if the product topology on $U\times V$ can come from a norm (which it can). This is different than what I ask.
 A: If $X$ is any infinite dimensional normed linear space there is a discontinuous linear functional $g$ on it. Let $x_0$ be a vector with $g(x_0)\neq 0$. Let $U$ be the kernel of $g$ and $W=span \{x_0\}$. Then $X=U+W$ (any $x\in X$ can be written as $(x-cx_0)+cx_0$ where $c=\frac {g(x)}{g(x_0)}$).
If we choose $(x_n)$ such that $x_n$ tends  to $0$ but $g(x_n)$ does not tend to $0$ then it is an easy matter to check that $f^{-1}((x_n-c_nx_0)+ c_n x_0)$ does not tend to $(0,0)$ so the inverse of your map $f$ is not continuous.
A: Consider an unbounded linear functional $u:X\to\Bbb{R}$ . Let $u(y)=1$ for some $y\in X$. Then define $P(x)=u(x)y$ . Then $P$ is not bounded as there exists a sequence of unit vectors $x_{n}$ such that $|u(x_{n})|\to\infty$ . Then $||P(x_{n})||\to \infty$
Then $P^{2}(x)=u(x)u(y)y=u(x)y=P(y)$ . Define $U=\ker(I-P)$ and let $V$ be such that $X=U\oplus V$. Then the product topology is the coarsest topology which makes $P$ and $I-P$ continuous . But $P$ is not continuous under the norm topology.
