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If $A,B$ are $\mathbb R$-Banach spaces, let $A\:\hat\otimes_\pi\:B$ denote the completion of the algebraic tensor product of $A$ and $B$ with respect to the projective norm. Let $X,Y,E,F$ be $\mathbb R$-Banach spaces. If $S\in\mathfrak L(X,E)$ and $T\in\mathfrak L(Y,F)$, let $S\otimes_\pi T$ denote the unique bounded linear operator from $X\:\hat\otimes_\pi\:Y$ to $E\:\hat\otimes_\pi\:F$ with $$(S\otimes_\pi T)(x\otimes y)=Sx\otimes Ty\;\;\;\text{for all }(x,y)\in X\times Y\tag1.$$ Note that $$\mathfrak L(X,E)\times\mathfrak L(Y,F)\to\mathfrak L(X\:\hat\otimes_\pi\:Y,E\:\hat\otimes_\pi\:F)\;,\;\;\;(S,T)\mapsto S\otimes_\pi T\tag2$$ is a bounded bilinear operator and hence there is a unique bounded linear operator $\iota$ from $\mathfrak L(X,E)\:\hat\otimes_\pi\:\mathfrak L(Y,F)$ to $\mathfrak L(X\:\hat\otimes_\pi\:Y,E\:\hat\otimes_\pi\:F)$ with $$\iota(S\otimes T)=S\otimes_\pi T\;\;\;\text{for all }(S,T)\in\mathfrak L(X,E)\times\mathfrak L(Y,F)\tag3.$$

Is $\iota$ injective?

I've seen that the notation $S\otimes T$ is often used for $S\otimes_\pi T$. With that notation it's not clear if it refers to the element of $\mathfrak L(X,E)\otimes\mathfrak L(Y,F)$ or the element in $\mathfrak L(X\:\hat\otimes_\pi\:Y,E\:\hat\otimes_\pi\:F)$. So, I guess they can be identified.

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Not in general. Consider the particular case when $E=F=\mathbb{R}$, then we are talking about the injectivity of the canonical map $$ X' \hat{\otimes}_\pi Y' \to \big( X \hat{\otimes}_\pi Y \big)'$$ where the $'$ denotes the dual space. However, this map factors as $$ X' \hat{\otimes}_\pi Y' \to X' \hat{\otimes}_\varepsilon Y' \to \big( X \hat{\otimes}_\pi Y \big)'$$ where the first arrow is simply the fact that $\varepsilon \le \pi$, and the second one can be found e.g. in Section 6.1 of Defant and Floret's Tensor norms and operator ideals. If the composition is injective then the first map $$ X' \hat{\otimes}_\pi Y' \to X' \hat{\otimes}_\varepsilon Y' $$ has to be injective, but this is well-known not to be the case in general (and this is closely related to approximation properties of the spaces involved; see Section 5 in the aforementioned book).

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    $\begingroup$ But, at least for your choice $E=F=\mathbb R$, it's correct if $Y$ has the approximation property, right? If so, the claim remain correct for more general $E,F$? $\endgroup$ – 0xbadf00d May 4 '18 at 23:10

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