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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?

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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).

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    $\begingroup$ 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. $\endgroup$ Commented Jul 22, 2022 at 20:17
  • $\begingroup$ 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. $\endgroup$ Commented Jul 22, 2022 at 22:08
  • $\begingroup$ 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 $U_i \subseteq U$. $\endgroup$ Commented Jul 22, 2022 at 22:10
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    $\begingroup$ ... 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). $\endgroup$ Commented Jul 22, 2022 at 22:16

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