# How can we show that $E\:\hat⊗_π F→E\:\hat⊗_ε\:F↪\mathfrak B(E',F')$ is injective iff $E\:\hat⊗_π F→(E'⊗_εF')'↪\mathfrak B(E',F')$ is injective?

In Tensor Norms and Operator Ideals by Defant and Floret, the authors state in Remark 5.4 that the "natural" mappings $$E\:\hat\otimes_\pi F\to E\:\hat\otimes_\varepsilon\:F\hookrightarrow\mathfrak B(E',F')\tag1$$ $$E\:\hat\otimes_\pi F\to (E'\otimes_\varepsilon F')'\hookrightarrow\mathfrak B(E',F')\tag2$$ are both injective, if one of them is. Above, $E,F$ are Banach spaces, $\hat\otimes_\pi$ denotes the completion wrt projective tensor norm, $\hat\otimes_\varepsilon$ denotes the completion wrt injective tensor norm, $\otimes_\varepsilon$ denote the injective tensor product, $\mathfrak B(E',F')$ denotes the space of bounded bilinear forms on $E'\times F'$ and $\hookrightarrow$ indicates an injection.

I don't get why this is correct. How can we show it?

(Is this just a simple result which is correct for any mappings $A\to B\hookrightarrow C$ and $A\to\tilde B\hookrightarrow C$?)

Isn't this just logic together with the simple fact that a mapping $f:X\to Y$ is injective if and only if it left-cancelable (i.e. for all $h,i:E\to X$ with $f\circ h=f\circ i$ one has $h=i$)?
Let $f:A\to B$, $\tilde f:A\to \tilde B$, $g:B\hookrightarrow C$ (i.e., injective) and $\tilde g:B\hookrightarrow C$ with $g\circ f=\tilde g\circ\tilde f$ be given and assume that $\tilde f$ is injective. Given $h,i :E\to A$ with $f\circ h= f\circ i$ we get $\tilde g\circ \tilde f\circ h= g\circ f \circ h= g\circ f \circ i = \tilde g\circ \tilde f\circ i$ and this implies $h=i$ because $\tilde g\circ \tilde f$ is injective.