# >Show that $\overline{T}$ is open if and only if $T$ is open. Furthermore, if $T$ us continuos, then $\| \overline{T}\|=\| T\|$.

Let $X,Y$ normed space, $T:X\to Y$ a linear map and $M$ a vector closed subspace of $X$ such that $M\subset\mbox{Ker}(T)$. Let $\overline{T}:X/M\to Y$ an application such that $T=\overline{T}\circ\omega$, where $\omega:X\to X/M$ a canonical map in the quotient.

Show that $\overline{T}$ is open if and only if $T$ is open. Furthermore, if $T$ us continuos, then $\| \overline{T}\|=\| T\|$.

My attemp: I know that, if $T:X\to Y$ is a linear map and $M$ a subvectorial space of $X$ such that $M\subseteq \mbox{Ker}(T)$, then there exist a unique linear map $\overline{T}=X/M\to Y$ such that $T=\overline{T}\circ\omega$, where $\omega$ is the canonical map in the quotient, and furthermore, $$\{ T(x)\mid x\in X\}=\{\overline{T}(x)\mid x\in X\}$$

So, I need to prove this map is open, for this I try to use the open map theorem: Let $X,Y$ banach space and let $T:X\to Y$ a linear continuos and surjective map, then $T$ is an open map. But in this problem I have a normed space $X, Y$ and I should prove this is an surjective map. And I don't see how prove the norm are equal. Thanks!

$|T(x)|=|\overline{T}(\omega(x))|\leq\|\overline{T}\|\|\omega(x)\|\leq\|\overline{T}\|\|x\|$, so $\|T\|\leq\|\overline{T}\|$.
$|\overline{T}(x+M)|=|\overline{T}(\omega(x))|=|T(x)|\leq\|T\|\|x\|$, so $\|\overline{T}\|\leq\|T\|$.
Note that $\|\omega(x)\|=\|x+M\|=\inf\{\|x+z\|: z\in M\}\leq\|x\|$.
For the openness, try to show that $\omega(B_{X})=B_{X/M}$, here $B$ stands for open unit ball.