I am dealing with the following exercise:
Let $E$, $F$ be Banach Spaces and $T:E\to F$ a continuous and surjective linear function. Define $\widehat{T}:E/\ker T\to F$ as $\widehat{T}([x])=T(x)$. Prove that $\widehat{T}$ is an isomorphism such that $\left \|T\right \|=\left \|\widehat{T}\right \|$.
First of all, since $T$ is continuous then $\ker T$ is closed, so $E/\ker T$ is a Banach space (because $E$ is a Banach Space). Take $Q:E\to E/\ker T$ the projection. Therefore $\widehat{T}\circ Q=T$, so $\widehat{T}$ is a continuous linear function which is bijective. By the open mapping theorem, it is also an homeomorphism. Moreover, $\left \|T\right \|=\left \|\widehat{T}\circ Q\right \|\leq \left \|\widehat{T}\right \|\left \|Q\right \|\leq\left \|\widehat{T}\right \|$ since $\left \|Q\right \|\leq 1$.
The only thing I need to prove in order to finish the exercise is the other inequality: $\left \|\widehat{T}\right \|\leq \left \|T\right \|$. I do not know how to achieve it. How would you solve this last step?
Pick some class $[x] \in E/ker(T)$ and an arbitrary $y \in [x]$. Then $$ \| \hat{T}([x]) \|= \|T(y)\| \leq \|T \| \| y \| $$
Therefore, $$\| \hat{T}([x]) \| \leq \|T \| \inf_{y \in [x]} \{ \|y \| \} =\|T \| \bigl\| [x ]\bigr\|$$
Added If you want to avoid the definition with inf, here is how you can re-write the proof:
Let $x \in E$ and $z \in \ker(T)$. Then
$$ \| \hat{T}([x]) \|= \|T(x-z)\| \leq \|T \| \| x-z \| = \| T \| d(x,z) $$
Therefore $$ \| \hat{T}([x]) \| \leq \| T \| \inf_{z \in \ker(T)} \{ d(x,z) \}= \|T \| \| [x]\| $$