Existence of the restriction of a global section on a sheaf.

Question

Let $\mathcal{A}$ be a sheaf of groups on $X$

An $\mathcal{A}$-sheaf on $X$ is a sheaf of sets $\mathcal{F}$ on $X$ together with a morphisms of sheaves of sets $$ a: \mathcal{A} \times \mathcal{F} \rightarrow \mathcal{F} $$ such that for every open subset $U \subseteq X$ the map $$ a_U: \mathcal{A}(U) \times \mathcal{F}(U) \rightarrow \mathcal{F}(U) $$ is an action of the group $\mathcal{A}(U)$ on the set $\mathcal{F}(U)$.

A morphism $\mathcal{F} \rightarrow \mathcal{G}$ of $\mathcal{A}$-sheaves is a morphism $\varphi: \mathcal{F} \rightarrow \mathcal{G}$ of sheaves such that for all $U \subseteq X$ the map $$ \varphi_U: \mathcal{F}(U) \rightarrow \mathcal{G}(U) $$ is $\mathcal{A}(U)$-equivariant. We obtain the category of $\mathcal{A}$-sheaves on $X$.

An $\mathcal{A}$-sheaf $T$ on $X$ is called an $\mathcal{A}$-pseudotorsor if for all $U \subseteq X$ open the action $\mathcal{A}(U) \times T(U) \rightarrow \mathcal{T}(U)$ is simply transitive.

An $\mathcal{A}$-pseudotorsor $T$ on $X$ is called an $\mathcal{A}$-torsor if there exists an open covering $X=\bigcup_i U_i$ such that $T\left(U_i\right) \neq \emptyset$ for all $i$. We obtain the full subcategory of $\mathcal{A}$-torsors of the category of $\mathcal{A}$-sheaves. We denote this category by $(\operatorname{Tors}(\mathcal{A}))$.

Problem: An $\mathcal{A}$-torsor $T$ over $X$ is trivial if and only if $T(X) \neq \emptyset$.

Proof If $T \cong \mathcal{A}$, then $T(X) \neq \emptyset$ because $\mathcal{A}(X)$ is a group, in particular it is non-empty. Conversely, assume there exists $s \in T(X)$. Then $$ \mathcal{A} \rightarrow T, \quad \mathcal{A}(U) \ni g \mapsto g \cdot\left(s_{\mid U}\right) \in T(U), \quad U \subseteq X \text { open } $$ is an isomorphism of $\mathcal{A}$-torsors.

My question is why this $s|_U=:res_U^X(s)$ exists? How we are sure that $T(U)$ where $U\subset X$ is nonempty? In general definition of sheaves I don't see a contradiction where $T(U)$ can't be an empty set. But in the definition of torsor there seems also no restriction (by the way the whole idea is the existence of a cover that and we are only sure on each patch of the cover our sheaf of sets is not empty.)

Well, $T$ is a presheaf, so restriction always makes sense. If $T(U)$ is empty then $T(W)$ is empty for all open $W\supseteq U$, fine, but the presheaf axioms certainly say there is a function $T(X)\to T(U)$ — FShrike, May 07 '24 at 22:38
@FShrike yes, there is a function by axioms of presheaves, so by the following:https://math.stackexchange.com/questions/789123/why-is-there-no-function-with-a-nonempty-domain-and-an-empty-range
$T(U)$ is then guaranteed not to be empty then.... — Micheal Brain Hurts, May 07 '24 at 22:44
so the question is turned out to be a very fundamental question from basic set theory :D — Micheal Brain Hurts, May 07 '24 at 22:46

score 2 · Accepted Answer · answered May 07 '24 at 22:39

$T$ is by definition a sheaf on $X$, so the inclusion of the open set $U\subseteq X$ induces by definition a restriction map $\mathrm{res}^X_U\colon T(X)\rightarrow T(U)$. This is part of the data that defines a sheaf. In general, a sheaf can be empty over some open subset, but if $T(X)\neq\emptyset$, this forces $T(U)\neq\emptyset$ because this map exists.

Existence of the restriction of a global section on a sheaf.

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