Is there an operator $T:X\rightarrow Y$ that factors through a Banach space $Z$ containing no complemented copy of $l\_{1}$, but does not factor through any Banach space $W$ containg no copy of $l\_{1}$? 



An operator factoring through a Banach space containing no copy of $l_{1}$
This is an explanation of Johnson's answer. We need to prove two things: (1) An identity on $C(0,1)$ does not factor through space which does not contain $\ell\_1$. (2) $C(0,1)$ does not contain a complemented copy of $\ell\_1$ (of course the identity on $C(0,1)$ factors through $C(0,1)$). 


(1) By the Banach-Mazur theorem $C(0,1)$ contains an isometric copy of $\ell\_1$, let $\{e\_i\}$ be its unit vector basis. Let $I=AB$ be the factorization through a Banach space $W$. We can pick $A^{-1}e\_i$ to be uniformly bounded, it is easy to check that they span a subspace isomorphic to $\ell\_1$ in $W$.


(2) Can be derived from the theory of $\mathcal{L}\_p$ spaces. Any complemented subspace of $C(0,1)$ is an $\mathcal{L}\_\infty$ space (See Lindenstrauss-Tzafriri Classical Banach spaces 1973), and thus contains $\{\ell\_\infty^n\}\_{n=1}^\infty$ with uniformly bounded distortions. The space $\ell\_1$ does not satisfy this condition as it has cotype 2.


P.S. An easier way to show (2) which goes back to Banach and Mazur, is to observe that otherwise the dual space of $C(0,1)$, which is weakly sequentially complete would contain an isomorphic copy of $\ell\_\infty$, which is not weakly sequentially complete (this is already in the comments of Chen).


