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Question: Does there exist an isomorphic predual of $\ell^1$, which does not have a quotient isomorphic to $c_0$?

Thanks in advance.

Edit: The answer is no. Let $X$ be a Banach space such that $X^*$ is isomorphic to $\ell^1$. If there didn't exist any surjective bounded $T:X\to c_0$, then $X$ would be a Grothendieck space. Being separable, $X$ would be reflexive.

It is awkward to answer your own question in a few hours after you post it. Please accept my apologies, I will delete this post in a couple of days.

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In my weak$^*$ basic sequences paper with Rosenthal we proved that if $\ell_1$ embeds into $X^*$ and $X$ is separable, then $c_0$ is isomorphic to a quotient of $X$.

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  • $\begingroup$ Professor Johnson, thank you truly for your answer. I must admit I hadn't read your paper you've referenced carefully enough. To redeem my shortcoming: I believe the first corollary in your 1977 paper with Hagler imply that any predual $X$ of $\ell^1(\Gamma)$ for an arbitrary set $\Gamma$ has quotients isomorphic to $c_0$, for $\ell^1(\Gamma)$ has RNP so $X$ cannot contain a copy of $\ell^1$. $\endgroup$
    – Onur Oktay
    Commented Jan 31, 2023 at 20:15
  • $\begingroup$ Correct, although this follows formally from the Josephson-Nissenzwieg theorem and my paper with Rosenthal. $\endgroup$
    – Bill Johnson
    Commented Jan 31, 2023 at 22:37

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