Analogy of fundamental theorem of homomorphism of Banach space I would like to prove the fundamental theorem of homomorphisms of Banach space.
Let $V$ and $W$ be Banach spaces.Let $f:V→W$ be surjective bounded linear map.
I would like to prove $f-:V/\ker f→W$ is continuous.
I understood $\ker f$ is closed in $V$, so $V/\ker f$ is a Banach space.
If I could prove this, I would be able to prove a kind of analogy of the fundamental theorem of homomorphisms for Banach spaces. Thank you in advance.
 A: There is a more general result which can help:

Let $V$ and $W$ be normed spaces, $V_0 \subset V$ is a closed vector
subspace, $T: V \to W$ is a bounded (i.e. continuous) linear operator,
$V_0 \subset \ker T$. Then:

*

*There's a surjective linear operator $Q: V \to V/V_0$ uniquely defined by $Q(x) = x + V_0$ $\forall x \in V$. This operator is called the quotient map of $V/V_0$. It is bounded, $||Q|| = 1$.

*There exists a unique bounded linear map $S : V/V_0 \to W$ s.t. $\forall x \in V$ $Tx = (SQ)x$. Moreover, $||S|| = ||T||$.


Proof
The surjectivity and linearity of $Q$ are obvious. Denote $\mathbb B_V$ and $\mathbb B_{V/V_0}$ the open unit balls in $V$ and $V/V_0$ respectively. We have the obvious inequality:
$$
||Q|| = \sup_{x \mathbb B_V}||Qx||_{V/V_0} = \sup_{x \in \mathbb B_V}||x + V_0||_{V/V_0} \leq \sup_{x \in \mathbb B_V}||x||_V = 1,
$$
proving $Q$ is bounded and $Q(\mathbb B_V) \subset \mathbb B_{V/V_0}$.
On the other hand, $\forall y \in \mathbb B_{V/V_0}$ $\exists x \in Q^{-1}(y) \subset V$ s.t. $||x||_V < 1$, because $||y||_{V/V_0} = \inf_{x \in Q^{-1}(y)}||x||_V < 1$. Hence, $Q(\mathbb B_V) \supset \mathbb B_{V/V_0}$.
Therefore, $Q(\mathbb B_V) = \mathbb B_{V/V_0}$. This clearly implies $||Q|| = 1$.
We already know that there exists a unique linear $S : V/V_0 \to W$ s.t. $\forall x \in V$ $Tx = (SQ)x$. This follows from a basic linear algebra.
Since $Q(\mathbb B_V) = \mathbb B_{V/V_0}$, $S(\mathbb B_{V/V_0}) = T(\mathbb B_V)$. Hence $S$ is bounded, and $||S|| = ||T||$. Q.E.D.
You can apply this statement to $f: V \to W$ and the quotient $V/\ker f$. Then you will get $\hat f : V/\ker f \to W$. It is surjective because $f$ is surjective, and injective by the construction. So $\hat f$ is a continuous bijection between Banach spaces. By the inverse mapping theorem it has a continuous inverse $\hat f^{-1}$. Therefore, $W$ and $V/\ker f$ are topologically isomorphic.
A: For all $x \in Ker f$, $u\in V$ you have:
$$\| \overline{f}(\overline{u})\|_W=\| f(u+x)\|_W \le \|f\|_{L(V,W)}\|u+x\|_V$$
So
$$\| \overline{f}(\overline{u})\|_W \le \|f\|_{L(V,W)} \inf_{x \in Kerf}\|u+x\|_V= \|f \|_{L(V,W)} \| \overline{u}\|_{V/kerf} $$
So...
