
> 
> **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.



$\ell^1$ predual with no $c_0$ quotient?
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 $X$.


