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Martin Bradendenburg proved here that if we have two maps between vector spaces $T\colon X\rightarrow Y,\;S\colon X'\rightarrow Y'$ then kernel of $T\otimes S$ is given by $\ker(T\otimes S)=\ker(T)\otimes X'+X\otimes\ker(S)$.

Is similar formula true in the setting of Banach spaces? To be more precise: assume now that $X,X',Y,Y'$ are Banach spaces, $T,S$ bounded operators as above and $X\otimes X',\,Y\otimes Y'$ stands for completion of an algebraic tensor product with respect to some cross norm (in particular I am interested in the case of $C^*$ algebras and minimal tensor product). Now, is

$\ker(T\otimes S)=\overline{\ker(T)\otimes X'+X\otimes\ker(S)}$

true?

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  • $\begingroup$ It's certainly true that $\overline{\ker(T)\otimes X'+X\otimes\ker(S)} \subseteq \ker (T \otimes S)$, and I'm inclined to think that the reverse inclusion holds but the proof doesn't pop into my head right away. $\endgroup$ Commented Nov 8, 2017 at 21:01

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Alright, i know the answer, in case anyone is interested i will post it in here.

The answer is that equality $\ker(T\otimes S)=\overline{\ker(T)\otimes X'+X\otimes\ker(S)}$ does not hold in general, it even does not need to hold when maps are $\ast$-homomorphisms between $C^*$-algebras (and we use minimal tensor product). It follows from the fact that not every $C^*$-algebra is exact (look here https://en.wikipedia.org/wiki/Exact_C*-algebra for definition of exactness of $C^*$-algebras). Take non-exact $C^*$-algebra, short exact sequence that gets tensored to non-exact sequence and see that non-exactness of this sequence forces above equality to be untrue for one of the maps.

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