TITLE: Quotients of l^infty
QUESTION [10 upvotes]: Let $M$ be a closed subspace of $l^\infty$. Suppose that the quotient $l^{\infty}/M$ is isomorphic to $l^\infty$. Is it true that $M$ is complemented in $l^\infty$?

REPLY [10 votes]: Notice that the comment by Yemon gives a negative answer to the question. Bourgain constructs a short exact sequence 
$0 \to \ell_1 \to \ell_1 \to X \to 0$
that does not split.  The space $X$ is not $\mathcal{L}_1$ for then (since the kernel of the quotient map is a dual space and the quotient mapping onto any $\mathcal{L}_1$ space locally lifts) the quotient map would lift by a classical result of J. Lindenstrauss. Dualizing the diagram, you get a quotient mapping from $\ell_\infty$ onto $\ell_\infty$ whose kernel is not $\mathcal{L}_\infty$ and thus is not complemented in $\ell_\infty$.  
Incidentally, from such an $\ell_\infty$ example you can deduce the $\ell_1$ result by localization; that is, modulo standard localization arguments, your question is equivalent to the question Bourgain answered. The moral is "don't look for an easy proof".