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.

**EDIT**: If the claim is not true in this generality, feel free to add additional assumptions (like the approximation property) on the involved spaces.