# Are all morphisms of sheaves of sections of "TVS bundles" obtained from TVS bundle morphisms? Even in the case of infinite rank?

Let $$X$$ be a topological space. By a morphism of base-$$X$$ sheaves, we mean a morphism of sheaves on $$X$$, whose underlying base space map is the identity on $$X$$.

We will use a stronger notion of a vector bundle, which we refer to as a "TVS bundle": this is a vector bundle with a topological vector space (TVS) structure on each fiber, and the "typical fiber" (used in the trivializations) is a TVS, and the local trivializations are, when restricted to each fiber, continuous linear with continuous inverse.

Suppose $$F,F'$$ are sheaves of real vector spaces on $$X$$, and suppose each of $$F,F'$$ is the sheaf of sections of some TVS bundle $$E,E'$$ (not necessarily of finite rank) on $$X$$.

My question is, does every morphism $$F \rightarrow F'$$ of base-$$X$$ sheaves (of real vector spaces) arise from a morphism $$E \rightarrow E'$$ of TVS bundles over $$X$$ (lying over the identity on $$X$$)? If yes: what tools are used to show this; and, is the morphism $$E \rightarrow E'$$ uniquely determined?

Do you mean for $$F$$ and $$F'$$ to be sheaves of topological real vector spaces on $$X$$? Otherwise the answer is no for trivial reasons:
Take $$X$$ to be the one-point space. Let $$E = E'$$ be an infinite-dimensional normed vector space, viewed as a TVS bundle on $$X$$. Let $$f : E \to E'$$ be any linear function which is not continuous. Then $$F = F'$$ is the underlying vector space of $$E = E'$$ (viewed as a sheaf of vector spaces on $$X$$) and $$f : F \to F'$$ is a well-defined morphism of sheaves, but $$f$$ does not arise from a morphism of TVS bundles $$E \to E'$$ (since it is not continuous).
• I see, thank you! Yes, I had a suspicion that we (at least) need to make $F,F'$ sheaves of TVSs, and require only sheaf morphisms which are TVS maps on each open of $X$. Thank you for explaining this. However, I'm not sure what kind of TVS structure should be used for the sheaf of sections of a TVS bundle. Commented Jul 22, 2022 at 20:17
• I think the topology of $F(U)$ ($F$ is the sheaf of sections of a TVS bundle $E$ with fiber $V$) should be constructed something like this: pick an open cover $\{U_i\}_i$ of $U$ and trivializations $\{\psi_i : E|_{U_i} \to U_i \times V\}_i$. Also pick open subsets $\{V_i\}_i$ of $V$ with all but finitely many equalling $V$ itself. Then $\{\sigma \in F(U) : \forall i \forall x \in U_i \; \psi_i(\sigma(x)) \in U_i \times V_i\}$ should be an open subset of $F(U)$, and the topology on $F(U)$ should be generated by these sets. Commented Jul 22, 2022 at 22:08
• In other words, $F(U)$ should have the coarsest topology making the canonical map $F(U) \to F(U_i) \to V$ continuous for all trivializations of $E$ over open subsets $U_i \subseteq U$. Commented Jul 22, 2022 at 22:10
• ... oops sorry I mean the canonical map $F(U) \to F(U_i) \to \operatorname{Hom}(U_i, V)$, with the topology of pointwise convergence (aka the product topology) on $\operatorname{Hom}(U_i, V)$. But another choice might be to use the compact-open topology; I'm not totally sure what's best (nor have I actually checked that this forms a sheaf). Commented Jul 22, 2022 at 22:16