# Exactness of injective tensor products

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.

• Do you mean just the injective tensor product or the completed injective tensor product? Jun 30, 2020 at 9:59
• @JochenWengenroth (I made the correction) but indeed I'm interested in the completed injective tensor product. Jun 30, 2020 at 10:05

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.