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Let $X, Y$ be two Banach spaces and $T:X\to Y$ a continuous surjection. Assume $Z$ is a complemented subspace of $X$ and that $T(Z)$ is closed. Is it necessarily true that $T(Z)$ is complemented in $Y$? I am rather convinced that this cannot be true, but I can't find a counterexample (though admittedly, Banach geometry is not really my area of expertise)
A few of my observations:
- To find a counterexample, $X$ and $Y$ cannot be Hilbert
- If we do not require $T(Z)$ to be closed, $T(Z)$ does not have to be complemented, and this can happen even in the $X,Y$ Hilbert case (example: $\varphi:\ell^2\to \ell^2, \varphi(\sum x_n e_n):=\sum x_{2n}e_{n}$ sends the closed subspace $Z:=\{x_{2n}=\frac{x_{2n-1}}{2n}\}$ to a dense subspace of $\ell^2$ which does not equal $\ell^2$).
- One possible idea would be to try and construct a right inverse $S$ (if it exists) such that $TS=Id_Y$ and then try to push the projection $P:X\to Z$ to one on $Y$ through $Q:=TPS$, though I can only prove that this is indeed a projection only with the additional assumption of $ST=\text{Id}_Z$, which does not look particularly wieldy (the only "natural" case in which this happens that I can think of is that of an $T$ being an isomorphism)
- To construct a counterexample, I was thinking of using the fact that each separable Banach space $X$ there is a surjection $P:\ell^1\to X$: if we take $X$ a separable H.I. space then it suffices to find a subspace $S$ of $\ell^1$ such that $P(S)$ is closed: in other words, I need $S+\text{ker}(T)$ to be closed. I am fairly sure this can be arranged but I can't prove it right now so any suggestion/comment/reference would be very welcome
