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Assume that $X$ and $Y$ are two Banach spaces and $T:X\to Y$ is a bounded surjective linear operator.

A consequence of the Michael selection theorem is that:"There is a continuous function $g:Y\to X$ such that $T\circ g=Id_{Y}$".

Can we always find a linear map $g$ as above?

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Would the following be an answer? There is no bounded linear 1-1 mapping $\ell^\infty/c_0 \to \ell^\infty$ Corollary 20 in link – Mirko Mar 23 '14 at 0:22
The title is not very apt: the previous comment explains why. Your question is equivalent to: "is a closed subspace of a Banach space always complemented?" and the answer is "usually no". – Yemon Choi Mar 23 '14 at 0:27
There are always a continuous selection and linear selection (playing with Hamel bases) but of course not always a continuous linear one. – Jochen Wengenroth Mar 24 '14 at 7:33
up vote 2 down vote accepted

Answer No. Just to add that the example that there is no bounded linear 1-1 mapping $\ell^\infty/c_0 \to \ell^\infty$ is due to Phillips in 1940 (as commented in the link provided by Yemon Choi above, discussing the complementary subspace problem, also credited to Phillips in the link to pdf I found and posted).

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