I've seen claims that it is known that for a pair of bounded injective linear operators $T\colon X\to Y, S\colon W\to V$, their tensor product $T\otimes S\colon X \otimes\_\pi W\to Y \otimes\_\pi V$ need not be injective. Here $\otimes\_\pi$ stands for the projective tensor product of Banach spaces.


1. Can this happen when $T = {\rm id}\_X$, the identity operator on some Banach space $X$?
2. If so, can it happen for $T = {\rm id}\_{L\_1}$, the identity operator on $L\_1$?


Question 2 has negative answer when $S$ is an isomorphism onto its range. Not surprisingly, the answer would be always [positive for the injective tensor product](https://mathoverflow.net/questions/364504/exactness-of-injective-tensor-products?rq=1).



Projective tensor product of injective operators
$\require{AMScd}\newcommand{\id}{\operatorname{id}}$I use a common characterisation of the [approximation property](https://en.wikipedia.org/wiki/Approximation_property) as found in e.g. Ryan's book [Zbl 1090.46001](https://www.zbmath.org/?q=an%3A1090.46001).



> 
> A Banach space $X$ has the approximation property if and only if for each Banach space $Y$ (it is enough to take $Y=X^\*$) the natural map
> $$ X \widehat\otimes Y \rightarrow X \check\otimes Y $$
> is injective.
> 
> 
> 


Here I write $\widehat\otimes$ and $\check\otimes$ for the completed projective, respectively, injective tensor products.


We can now answer (2) in the negative. Let $X$ have the approximation property, and let $S:W\rightarrow V$ be injective. Consider the commutative diagram
$$ \begin{CD}
X\widehat\otimes W @>>> X \check\otimes W \\
@V{\id\otimes S}VV @VV{\id\otimes S}V \\
X\widehat\otimes V @>>> X \check\otimes V
\end{CD} $$
The map $\id\otimes S: X \check\otimes W \rightarrow X \check\otimes V$ is injective, and the horizontal arrows are injective as $X$ has AP, so $\id\otimes S: X \widehat\otimes W \rightarrow X \widehat\otimes V$ is injective. In particular $X=L\_1$ has the AP, showing the negation of (2).


As Jochen Wengenroth noted, Q1 can be reduced to the $T\otimes S$ case which the OP stated has a positive answer. However, here is a concrete example, following Chapter 5, Corollary 4 of Defant and Floret [Zbl 0774.46018](https://www.zbmath.org/?q=an%3A0774.46018). Let $X$ be any Banach space, and let $B\_{X^\*}$ be the unit ball of the dual space $X^\*$, consider $\ell\_\infty(B\_{X^\*})$ and define $j:X\rightarrow \ell\_\infty(B\_{X^\*})$ by evaluation: $j(x) = ( \phi(x) )\_{\phi\in B\_{X^\*}}$. Then $j$ is an isometry onto its range. We know that $\ell\_\infty(B\_{X^\*})$ has AP so
$$ X^\* \widehat\otimes \ell\_\infty(B\_{X^\*}) \rightarrow X^\* \check\otimes \ell\_\infty(B\_{X^\*}) $$
is injective. Consider now the commutative diagram
$$ \begin{CD}
X^\* \widehat\otimes X @>>> X^\* \check\otimes X \\
@V{\id\otimes j}VV @VV{\id\otimes j}V \\
X^\* \widehat\otimes \ell\_\infty(B\_{X^\*}) @>>> X^\* \check\otimes \ell\_\infty(B\_{X^\*}) \\
\end{CD} $$
The bottom arrow is injective, and the right-hand down arrow is. If $X$ does not have AP then the top arrow is not injective, and so the left-hand
down arrow must fail to be injective, which gives an example of (1). (There is nothing special about $\ell\_\infty$ here: any Banach space $F$ with the AP and any injection $j:X\rightarrow F$ would work.)


