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The Banach space $C[0,1]$ is universal for all separable Banach spaces in the sense that for a separable Banach space $X$ there is an isometric isomorphism from $X$ into $C[0,1]$. My question is whether there is a universal operator in the following sense:

Do there exist separable spaces $X$ and $Y$ and a bounded linear operator $U:X \to Y$ so that if $A:E\to F$ ($E,F$ are separable Banach spaces) then $A=BUC$ where $C:E\to X$ and $B:Y\to F$. My guess is that the answer is no.

The reason I think the answer is no is that Johnson and Szankowski proved two results that somewhat contradict the affirmative: (1) There is no separable Banach space that is complementably universal for all separable Banach spaces (Studia 1976). (2) There is no separable Banach space that every compact operator factors through (JFA 2009).

If the answer is no to the first question then I'm interested in the same question for compact $A$.

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    $\begingroup$ If the identity operator on $E$ factors through $U:X\to Y$ via $(A,B)$, then $A$ is an isomorphic embedding and $BAU$ is a projection from $X$ onto $AE$, so you know answer to (1). Similarly, the answer to (2) is also negative because of JFA 2009. $\endgroup$
    – Bill Johnson
    Commented Jun 30, 2016 at 18:11
  • $\begingroup$ Bill: Thanks, should have seen that! $\endgroup$
    – Kevin Beanland
    Commented Jun 30, 2016 at 18:20
  • $\begingroup$ The problem would have seemed much better if I hadn't listed the Johnson-Szankowski papers. It took me longer to write than it should have taken me to come up with Bill's example once I found the J-S results. $\endgroup$
    – Kevin Beanland
    Commented Jun 30, 2016 at 19:32
  • $\begingroup$ It is worthwhile mentioning that these examples use the existence of many spaces that fail the bounded approximation property. They are needed because of the Kadec-Pelczynski example of a Banach space $X$ with a Schauder basis such that every separable Banach space with the bounded approximation property is isomorphic to a complemented subspace of $X$. $\endgroup$
    – Bill Johnson
    Commented Jul 1, 2016 at 11:48

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