Exactness of injective tensor products

Question

For (algebraic) tensor products, it is well-known that the functor $A\otimes_R \cdot:Mod_R\rightarrow Mod_R$ is only (left-) exact when $A$ is a flat $R$-module. In particular, all vector spaces are flat. What happens in the continuous (archimedean) setting?:

Let $B$ be a separable infinite-dimensional Banach space and suppose that $ f:E\rightarrow F, $ is a continuous linear injective map from a separable nuclear space $E$ to a separable Banach space $F$, both infinite-dimensional (if it matters). Let $\otimes_{\epsilon}$ denote the injective tensor product of LCS and let $\hat{\otimes}_{\epsilon}$ denote its completion.

Is the map $ 1_{B}\hat{\otimes}_{\epsilon} f: B\hat{\otimes}_{\epsilon} E \rightarrow B\hat{\otimes}_{\epsilon} F, $ a continuous linear 1-1 map also?

Related: This post is related to this unanswered post.

Answer 1

This is always true (without nuclearity): If $T_j:E_j\to F_j$ are continuous linear maps between Hausdorff locally convex spaces and $E_2$ is complete then $$ T_1\hat\otimes_\varepsilon T_2: E_1 \hat\otimes_\varepsilon E_2 \to F_1\hat\otimes_\varepsilon F_2$$ is injective if so are $T_1$ and $T_2$. This is 16.2.2 in Jarchows's book Locally Convex Spaces.