Topological direct sum of vector spaces Here is the theorem I want to prove (from wikipedia fr).
Let $V$ be a normed topological vector space and let $V_1$ and $V_2$ be two subspaces such that $V=V_1\oplus V_2$.
The sets $V_1$ and $V_2$ are equipped with the induced topology.
I suppose that the bijection $\psi: V_1\times V_2\to V$ given by $\psi(v_1,v_2)=v_1+v_2$ is homeomorphic (bijective, continuous wit continuous inverse).
Then:

*

*$V_1$ and $V_2$ are closed in $V$

*the projection $s:V_2\to V/V_1$ is homeomorphic.

I'm able to prove that $V_1$ and $V_2$ are closed and that $s$ is continuous.
I'm stuck on proving that $s^{-1}$ is continuous.
QUESTION: how to prove the continuity of $s^{-1}$ ?
Here is what I do. Let $O$ be open in $V_2$. There exists an open part $A$ of $V$ such that $O=A\cap V_2$.
Let $p:V\to V/V_1$ be the projection.
I know that $s(O)$ is open iff the part $p^{-1}(s(O))$ is open in $V$. I also knwo that
$p^{-1}(s(O))=O+V_1$
So let $x\in O+V_1$. I have to found a neighborhood of $x$ in $V$ which is contained in $O+V_1$. For the sake of simplicity I suppose $x\in O$.
from here, I'm quite sure that I'm going the wrong way
Let $r>0$ be such that $B(x,r)\subset A$. I want to prove that the ball $B(x,r)$ is included in $O+V_1$.
Let $y\in B(x,r)$. I write $y=x+h$ and I decompose $h=h_1+h_2$ with $h_1\in V_1$ and $h_2\in V_2$.
I have $\|h\|=\|x-y\|\leq r$ and then $\|h_2\|\leq r$ in such a way that $x+h_2\in B(x,r)\subset O$. Then $y=x+h_2+h_1\in O+V_1$.
Where it's wrong : I cannot deduce $\|h_2\|\leq r$ because $V_1$ and $V_2$ are not othonormal.
If the dimension is finite, I can create a scalar product adapted to the decomposition $V_1\oplus V_2$ and then invoke the fact that all the norms are equivalent.
EDIT -- QUESTION Is the statement true ? Bourbaki seems to assume completeness of the spaces (hence Banach)
 A: We can consider $V_1 \times V_2$ as direct sum $V_1 \oplus V_2$ with norm $$ \|(v1,v2) \| = \max\{\|v_1\|,\|v_2\|\} $$
Looking this way, $\phi$ and $\phi^{-1}$ are bounded linear transformations. Hence, there exists some $C>0$ such that $$ \|\phi^{-1}(v_1 + v_2)\| = \max\{\|v_1 \|,\|v_2\|\} \leq C \|v_1 + v_2\|$$
Now, let $x_n + V_1$ be a sequence in $\frac{V}{V_1}$ ( $x_n \in V_2$) such that
$$\| x_n + V_1\| = d(x_n, V_1) \rightarrow 0$$
where $d(x_n,V_1)$ is the distance between $x_n$ and $V_1$.
Hence, there exists a sequence $v_n \in V_1$ such that
$$ \|x_n + v_n\|\rightarrow 0$$
Hence $$ \| x_n \| \leq \max\{\|x_n \| , \|v_n \| \} \leq C\|x_n + v_n \|\rightarrow 0  $$
This proves that $s^{-1}$ is continuous at 0 and hence continuous.
A: The topology of $V$ does not need to be induced by a norm.
Since $\psi^{-1}:V\to V_1\times V_2$ continuous, its second component $q:V\to V_2$ is continuous.
But $q=s^{-1}\circ\pi$, where $\pi$ is the quotient map $V\to V/V_1$.
Hence $s^{-1}$ is continuous.
