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](https://en.wikipedia.org/wiki/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](https://mathoverflow.net/questions/86330/exactness-of-completed-tensor-product-of-nuclear-spaces).



Exactness of injective tensor products
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*.


