Is the image of a complemented subspace complemented?

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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

1 Answer

For a counterexample, let $$Y$$ be a Banach space with a closed, non-complemented subspace $$Z\subset Y$$. Consider the restriction of the sum operation $${\bf +}:Y\times Y\to Y$$ to the space $$X:=Y\times Z$$, and call it $$T:X\to Y$$.