Quotient of normed space ; Isometry I am struggling with the following problem:
$T : \mathbb{X} \to \mathbb{Y}$ is a linear operator, and $\mathbb{X}$ , $\mathbb{Y}$ are normed spaces.
Show that : $T$ is a "quotient map" iff the induced operator $T': \mathbb{X}/\ker(T) \to \mathbb{Y}$ is an isometry.
(I'm working with a German textbook and in that terminology a quotient map between normed spaces $\mathbb{X}$ and $\mathbb{Y}$ is a linear operator that maps the open unit ball in $\mathbb{X}$ surjectively onto the open unit ball in $\mathbb{Y}$ ; don't know whether that is a standard).
So far I could easily proof that if $T'$ is an isometry $T$ must be a "quotient map" , but I can't show the other implication, so I would appreciate a hint .
Thanks so far
Daniel

Suppose $T$ is such a quotient map, then we have $B_\mathbb{Y} \subset T(B_\mathbb{X})$ where $B_\mathbb{X}$ and $B_\mathbb{Y}$ are the open unit balls of $\mathbb{X}$ and $\mathbb{Y}$ respectively. Take any $x \in \mathbb{X}$ and let $y = Tx$. Without loss of generality we can assume $x \not \in \ker T$. Now for any $\epsilon \in (0,1)$, there's some $x_0 \in B_\mathbb{X}$ such that $Tx_0 = \epsilon \frac{y}{\lVert y \rVert}$. Now we find that
$$
\lVert x + \ker T \rVert \leq \epsilon^{-1} \lVert y \rVert \lVert x_0 \rVert \leq \epsilon^{-1} \lVert Tx \rVert,
$$
because $x-\epsilon^{-1} \lVert y \rVert x_0 \in \ker T$. Since $\epsilon$ was arbitrary we must have $\lVert Tx \rVert \geq \lVert x+ \ker T \rVert$.
All that is left to prove is that $\lVert Tx \rVert \leq \lVert x+ \ker T \rVert$. Note that for any $\epsilon > 0$ and $x \in \mathbb{X}$ we know that 
$$
\lVert T\left( \frac{x}{\epsilon+\lVert x \rVert } \right) \rVert < 1,
$$
since $T$ is a quotient map. So because $\epsilon >0$ was arbitrary we have $\lVert Tx \rVert \leq \lVert x \rVert$. Note that this implies that for any $x \in \mathbb{X}$ and $y \in \ker T$ we have $\lVert T(x+y) \rVert = \lVert Tx \rVert \leq \lVert x+y \rVert$. Taking the infimum results in $\lVert Tx \rVert \leq \lVert x+ \ker T \rVert.$ This concludes the proof that $T'$ is an isometry.
