First Isomorphism Theorem for Banach Spaces involving isometry?

Question

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?

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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.

Answer 1

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.

Answer 2

No this is not true. Take for example $X=\mathbb{R}^2$ equipped with the euclidean norm, $Y=\mathbb{R}^2$ with the one norm $||(x,y)||_1=|x|+|y|$ and let $T$ be the identity. Then your claim doesn’t hold.