An operator $T\colon X\rightarrow Y$ is said to be strictly cosingular provided that for no infinite-dimensional Banach space $Z$ there exist surjective operators $R\colon X\rightarrow Z$ and $S\colon Y\rightarrow Z$ such that $R=ST$; Equivalently, there is no infinite-codimensional subspace $V$ of $Y$ such that $Q\_{V}T$ is surjective.
Similarly, $T$ is said to be $Z$-strictly cosingular if there is no operator $S:Y\rightarrow Z$ for which $ST$ is surjective.


Question. If $T\colon X\rightarrow L\_{1}[0,1]$ is $l\_{1}$-strictly cosingular, is $T$ strictly cosingular?


Thank you!



Strictly cosingular operators and $l_{1}$-strictly cosingular operators into $L_{1}[0,1]$
Yes, they are the same. Pełczyński proved that strictly singular, strictly $\ell\_1$-singular, strictly cosingular, and weakly compact operators on $L\_1$ are all the same. This is Theorem 1 in



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> A. Pełczyński, On strictly singular and strictly cosingular operators. II. Strictly singular and strictly cosingular operators in $L\_1(\mu)$-spaces, *Bull. Bad. Pol. Sci.* **13** (1965) 31–36.
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Clearly, every weakly compact operator is strictly $\ell\_1$-cosingular. Let us take an operator that is not weakly compact. Now, by the above-mentioned result, $T$ fixes a copy of $\ell\_1$. Every copy of $\ell\_1$ contains a further copy of $\ell\_1$ that is complemented in $L\_1$. Take a projection $P$ onto a complemented copy of $\ell\_1$ in the range of $T$. Then $PT$ is a surjection onto $\ell\_1$, so $T$ is not strictly $\ell\_1$-cosingular.




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Every weakly compact operator is strictly $\ell\_1$-(co)singular. Conversely, every strictly $\ell\_1$-(co)singular operator into $L\_1$ is weakly compact. 


Indeed, if $T\colon X\to L\_1$ is *not* weakly compact then there is a sequence $(x\_n)$ in $X$ without a weakly convergent subsequence such that $(Tx\_n)$ does not have a weakly convergent subsequence either. As $L\_1$ is weakly sequentially complete, by Rosenthal's $\ell\_1$-theorem, $(x\_n)$ contains a sequence equivalent to the unit vector basis of $\ell\_1$. (Indeed, bounded operators map weakly Cauchy sequences into weakly Cauchy sequences but weakly Cauchy sequences in $L\_1$ converge weakly.) Let us then suppose that $(x\_n)$ and $(Tx\_n)$ are equivalent to the standard basis of $\ell\_1$, that is $T$ fixes a copy of $\ell\_1$. By [Proposition 5.7.2](https://books.google.co.uk/books?id=UG6zDAAAQBAJ&pg=PA134&dq=contains%20a%20complemented%20subspace%20of%20L1%20of%20l1&hl=pl&sa=X&ved=0ahUKEwjshseXh5TaAhVCAsAKHd-IAAcQ6AEINDAB#v=onepage&q=subspaces%20of%20L1&f=false) from [Albiac-Kalton], we may suppose that this copy is complemented, so let $P$ be a projection onto it. Then $PT$ is a surjection onto a space isomorphic to $\ell\_1$.


