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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.

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  • $\begingroup$ bounded and linear? $\endgroup$ – C.Ding Jul 19 '18 at 9:19
  • $\begingroup$ @C.Ding bounded, onto and linear. $\endgroup$ – Idonknow Jul 19 '18 at 9:21
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    $\begingroup$ @FriederJäckel quotient norm on $X/ker(T).$ $\endgroup$ – Idonknow Jul 19 '18 at 9:21
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    $\begingroup$ $X/\ker T$ is isomorphic to $Y$ (via the induced map) in the category of Banachable spaces, but usually the induced map is not isometric. More, generically there won't be an isometric isomorphism $X/\ker T \to Y$. The situation mentioned in your edit is special. $\endgroup$ – Daniel Fischer Jul 19 '18 at 11:24
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    $\begingroup$ @Idonknow : you are right saying that there's something else to prove in that proof of the theorem, but the way you tried to prove what's missing doesn't work. You have to prove not that every quotient map is an isometry, but that the specific map they built is an isometry, a fact that they don't prove in their proof. However, you can prove this claim with the same technique they use in the first part of the proof. $\endgroup$ – Bob Jul 19 '18 at 12:14
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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.

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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.

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