Factor spaces isometric isomorphism proof

Question

Let $X,Y$ be normed spaces and $T\in \mathcal{L}(X,Y)$. I want to prove that if there is $\hat{T}\ \colon X/\text{ker } T\to Y$, $$T(x+\text{ker }T)=Tx$$ which is isometric isomorphism, then $T$ is surjective and for all $y\in Y$: $$\|y\|=\inf \{\|x\|\ \colon x\in X, \ Tx=y\}.$$ Any ideas on how to prove this?

Answer 1

Presumably, you mean that $\hat T: X/\ker T \to Y$ is an isometric isomorphism.

First, note that if $\hat T$ is an isomorphism, then $\hat T$ is surjective, which means that $T$ must be surjective.

Now, $\hat T$ is an isometric isomorphism. So, that means that for every $y$, there is an $\hat x \in X/\ker T$ such that $\hat T(x) = y$, and this $\hat x$ satisfies $\|\hat x\| = \|y\|$.

However, by definition, there is an $x \in X$ such that $$ \hat x = x + \ker T = \{x' \in X \mid x' - x \in \ker T\} = \{x' \in X \mid Tx' = Tx\} = \\ \{x' \in X \mid Tx' = y\} $$ and by definition, we have $$ \|\hat x\| = \|\hat x\|_{X/\ker T} = \inf\{\|x'\|:x' \in x + \ker T\} $$ Perhaps from here, you can see how to put the pieces together.