Question: Let $X$ and $Y$ be Banach spaces.
Assume that $T:X\to Y$ is a bounded onto linear operator.
Is it true that $X/ker(T)$ is isometric isomorphic to $Y?$
In other words, do we have first isomorphism theorem for Banach spaces involving isometry?
EDIT: Motivation of the question comes from the following: To prove that every separable Banach space is linearly isometric to a quotient space of $\ell_1,$ Fabian et al. , in Chapter $5,$ Theorem $5.1,$ prove that the map
$$T:\ell_1 \to X$$
defined by
$$T((\xi_n)_n) = \sum_{n=1}^\infty \xi_n x_n$$
where $\{x_n:n\in\mathbb{N}\}$ is a dense subset of the closed unit ball of $X,$ is a bounded onto linear operator.
Then they conclude that $X$ is linearly isometric to $\ell_1/T^{-1}(0).$
It seems to me that they are using the first isomorphism theorem to obtain last sentence.
Best Answer
Very easily false! Let $X=\mathbb R^{n}$ with the $l^{1}$ norm and $Y=\mathbb R^{n}$ with the usual norm and $T$ be the identity operator. $Y$ is a Hilbert space and $X$ is not, so there is no isometric isomorphism.