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Let $\mathcal{X}$ be a decomposable Banach space (i.e. a topological direct sum of infinite-dimensional subspaces, say $\mathcal{X}=\mathcal{A}\oplus\mathcal{B}$). Can one always obtain another decomposition $\mathcal{X}=\mathcal{U}\oplus\mathcal{V}$ (of infinite-dimensional closed subspaces $\mathcal{U}$ with $\mathcal{V}$) with a continuous linear map $T\colon \mathcal{U}\to\mathcal{V}$ that admits a continuous right or left inverse?

If any partial/conditional results or known reformulations exist, that will be most appreciated.

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  • $\begingroup$ Sure, you can always adjoin some finite-dimensional subspace of $V$ to $U$. $\endgroup$ Apr 25, 2021 at 15:04
  • $\begingroup$ @Tomasz Kania. Could I kindly ask that you expand on your suggestion. (Seems I should easily see it but I’m not getting it right away ). $\endgroup$
    – Jack L.
    Apr 25, 2021 at 15:13
  • $\begingroup$ Finite-dimensional subspaces are always complemented, so write $U = U_1 \oplus F_1$ with $F_1$ finite-dimensional and take $U^\prime = U_1$ and $V^\prime = V + F_1$. $\endgroup$ Apr 25, 2021 at 15:31
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    $\begingroup$ Tomek, the OP wants $U$ to contain a complemented copy of $V$ or $V$ to contain a complemented copy of $U$ and $X=U\oplus V$. $\endgroup$ Apr 25, 2021 at 17:54
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    $\begingroup$ To get an example, take two HI spaces $U$ and $V$ s.t. every bounded linear operator from $U$ to $V$ is strictly singular and apply the Edelstien-Wojtasczcyk theorem (see vol.1 of Lindenstrauss-Tzafriri Theorem 2.c.13) to $X := U \oplus V$, $\endgroup$ Apr 25, 2021 at 17:56

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