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I want to verify whether

Given $X,Y$ are Banach spaces , a continuous surjective linear operator $T:X\to Y$ has a continuous right inverse .

It is well known that every surjective function has a right inverse . Now if I apply open mapping to that right inverse operator , then the right inverse is bounded . hence continuous . Hence the statement is true .

Is my observation correct ? It will be helpful if someone checks the argument . Regards .

am_11235...
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    The statement is wrong. See the answer by Tomasz Kania here: https://math.stackexchange.com/questions/2301255/surjective-bounded-operator-in-banach-spaces-without-bounded-right-inverse?rq=1 – Kavi Rama Murthy Jan 17 '23 at 05:27
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    Do you require a "right inverse" to be linear? The "right inverse" that you apply the open mapping theorem to in the above argument need not be linear. See Terry Tao's answer to https://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics ("This is subtle. Zorn's lemma gives a linear right inverse; the open mapping theorem gives a bounded right inverse. But getting a right inverse that is simultaneously bounded and linear is not always possible!") – leslie townes Jan 17 '23 at 07:20
  • Ok, if I drop the linearity on right inverse, then is the hypothesis true? (There is some Bartle-Graves theorem, but the answer by Tomasz Kania in the other link disapproves it, so I don't understand.) – am_11235... Jan 17 '23 at 08:59

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