Let $X,Y$ be two Banach spaces, and $T: X \to Y$ be a continuous linear surjective map. Let $\mathrm{ker} T = \{x \in X\mid Tx = 0\}$. It's well known that $X/\mathrm{ker}T$ is isomorphic to $Y$. My question is, is it possible to get a closed subspace $Z \subset X$ such that the restriction of $T$ to $Z$, $T|_Z: Z \to Y$ is an isomorphism (injective and onto)?
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1You certainly assume that $T$ is linear. – Paul Frost May 29 '25 at 10:26
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1No. See Example of a closed subspace of a Banach space which is not complemented – Just a user May 29 '25 at 11:14
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@Justauser How does that answer the question? – Dean Miller May 29 '25 at 11:30
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@gaoqiang Perhaps you can consider editing this post to add some more context so that it can be reopened since you already have a nice answer for this question. Some ideas can be to mention why you are interested in the result or to sketch some previous attempt you had at the result. – Dean Miller Jun 23 '25 at 07:48
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To expand my comment as an answer as requested ...
In homological terms, this is asking whether every short exact sequence of Banach spaces splits:
$$0\rightarrow \ker T\rightarrow X\rightarrow Y \rightarrow 0$$
And the answer is no according to the old question.
To add some details, given a Banach space $X$ and a closed subspace $X'$, let $T:X\rightarrow X/X'=:Y$. If we can find $Z\subset X$ such that $T|_{Z}:Z\rightarrow Y$ is an isomorphism of Banach spaces, then in particular $Z$ is complete hence closed in $X$, and at least algebraically $X=X'\oplus Z$, an internal direct sum.
The external direct sum $X'\oplus Z$ is a Banach space (with norms such as $\|(a,b)\|_{X'\oplus Z}=\|a\|_{X'}+\|b\|_{Z}=\|a\|_X+\|b\|_X$). It has an obvious continuous bijective map to $X$, and by open mapping theorem, it must be an isomorphism. Hence $X\simeq X'\oplus Z$ not only algebraically but also topologically. Therefore, if $X'$ is not complemented, such a $Z$ doesn't exist.
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