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][1] 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][2].  


  [1]: https://en.wikipedia.org/wiki/Injective_tensor_product
  [2]: https://mathoverflow.net/questions/86330/exactness-of-completed-tensor-product-of-nuclear-spaces