About an application of the open mapping theorem I need some help proving this result. I know that it has relation with the open mapping theorem, but I can't see how. It  states the following:
Let $S:E_1\to E_2$ surjective and linear, and $p:E_1\to\frac{E_1}{Ker S}$.
Then, if $S$ is continuous, $\overline{S}:\frac{E_1}{Ker S}\to E_2$, defined as $\overline{S}(Ker S + x)=S(x)$ is also continuous.
Moreover, if $E_1$ and $E_2$ are Banach spaces, then $\overline{S}$ is an isomporphism.
Thanks a lot in advance for any help.
 A: I suppose we can assume the fact that $\overline{S}\colon E_1/\ker S \to E_2$ is a well-defined bijective linear mapping as known. That is pure linear algebra, and has nothing to do with the topologies.
Now suppose that $E_1$ and $E_2$ are normed spaces, and $S$ is continuous. Then $\ker S$ is a closed subspace of $E_1$, and hence $E_1/\ker S$ is a normed space with the norm
$$\lVert p(x)\rVert_Q := \inf \left\{ \lVert x - y\rVert : y \in \ker S \right\}.$$
Let $U_1$ denote the open unit ball in $E_1$, and $U$ the open unit ball in $E_1/\ker S$, that is, $U_1 = \{x \in E_1 : \lVert x\rVert < 1\}$ and $U = \{\xi \in E_1/\ker S : \lVert \xi\rVert_Q < 1\}$.
Then we have $p(U_1) = U$.
On the one hand, if $x\in U_1$, we have $\inf \{ \lVert x- y\rVert : y \in \ker S\} \leqslant \lVert x\rVert < 1$, so $\lVert p(x)\rVert_Q  < 1$, i.e. $p(x) \in U$, and hence $p(U_1) \subset U$. On the other hand, if $\xi = p(x) \in U$, hence $\lVert p(x)\rVert_Q < 1$, by definition of the quotient norm, there is an $y \in \ker S$ with $\lVert x-y\rVert < 1$, i.e. $x-y \in U_1$. But $p(x) = p(x-y)$, and hence $p(x) = p(x-y) \in p(U_1)$.
Since $S$ is continuous, there is a $C < \infty$ with $\lVert S(x)\rVert_{E_2} \leqslant C$ for all $x \in U_1$, and by the above, that means $\lVert \overline{S}(\xi)\rVert_{E_2} \leqslant C$ for all $\xi \in U$, hence $\overline{S}$ is continuous with $\lVert \overline{S}\rVert = \sup\limits_{\xi\in U} \lVert \overline{S}(\xi)\rVert_{E_2} \leqslant C$.
So then $\overline{S}$ is a continuous linear bijection between normed spaces.
If $E_1$ is a Banach space, so is its quotient $E_1/\ker S$, and if also $E_2$ is a Banach space, the open mapping theorem asserts that $S$ and $\overline{S}$ are open mappings, in particular $\overline{S}$ is a continuous and open bijection, hence a homeomorphism, so $\overline{S}^{-1}$ is continuous, and $\overline{S}$ is an isomorphism of Banach spaces.
