Projective tensor product of injective operators 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.

*

*Can this happen when $T = {\rm id}_X$, the identity operator on some Banach space $X$?

*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.
 A: $\require{AMScd}\newcommand{\id}{\operatorname{id}}$I use a common characterisation of the approximation property as found in e.g. Ryan's book Zbl 1090.46001.

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.  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.)
