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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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    $\begingroup$ 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 $\endgroup$
    – Mirko
    Mar 23, 2014 at 0:22
  • $\begingroup$ 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". $\endgroup$
    – Yemon Choi
    Mar 23, 2014 at 0:27
  • $\begingroup$ See math.stackexchange.com/questions/411119/… $\endgroup$
    – Yemon Choi
    Mar 23, 2014 at 0:33
  • $\begingroup$ There are always a continuous selection and linear selection (playing with Hamel bases) but of course not always a continuous linear one. $\endgroup$
    – Jochen Wengenroth
    Mar 24, 2014 at 7:33

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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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