## 20.4 First cohomology and torsors

Definition 20.4.1. Let $X$ be a topological space. Let $\mathcal{G}$ be a sheaf of (possibly non-commutative) groups on $X$. A torsor, or more precisely a $\mathcal{G}$-torsor, is a sheaf of sets $\mathcal{F}$ on $X$ endowed with an action $\mathcal{G} \times \mathcal{F} \to \mathcal{F}$ such that

1. whenever $\mathcal{F}(U)$ is nonempty the action $\mathcal{G}(U) \times \mathcal{F}(U) \to \mathcal{F}(U)$ is simply transitive, and

2. for every $x \in X$ the stalk $\mathcal{F}_ x$ is nonempty.

A morphism of $\mathcal{G}$-torsors $\mathcal{F} \to \mathcal{F}'$ is simply a morphism of sheaves of sets compatible with the $\mathcal{G}$-actions. The trivial $\mathcal{G}$-torsor is the sheaf $\mathcal{G}$ endowed with the obvious left $\mathcal{G}$-action.

It is clear that a morphism of torsors is automatically an isomorphism.

Lemma 20.4.2. Let $X$ be a topological space. Let $\mathcal{G}$ be a sheaf of (possibly non-commutative) groups on $X$. A $\mathcal{G}$-torsor $\mathcal{F}$ is trivial if and only if $\mathcal{F}(X) \not= \emptyset$.

Proof. Omitted. $\square$

Lemma 20.4.3. Let $X$ be a topological space. Let $\mathcal{H}$ be an abelian sheaf on $X$. There is a canonical bijection between the set of isomorphism classes of $\mathcal{H}$-torsors and $H^1(X, \mathcal{H})$.

Proof. Let $\mathcal{F}$ be a $\mathcal{H}$-torsor. Consider the free abelian sheaf $\mathbf{Z}[\mathcal{F}]$ on $\mathcal{F}$. It is the sheafification of the rule which associates to $U \subset X$ open the collection of finite formal sums $\sum n_ i[s_ i]$ with $n_ i \in \mathbf{Z}$ and $s_ i \in \mathcal{F}(U)$. There is a natural map

$\sigma : \mathbf{Z}[\mathcal{F}] \longrightarrow \underline{\mathbf{Z}}$

which to a local section $\sum n_ i[s_ i]$ associates $\sum n_ i$. The kernel of $\sigma$ is generated by the local section of the form $[s] - [s']$. There is a canonical map $a : \mathop{\mathrm{Ker}}(\sigma ) \to \mathcal{H}$ which maps $[s] - [s'] \mapsto h$ where $h$ is the local section of $\mathcal{H}$ such that $h \cdot s = s'$. Consider the pushout diagram

$\xymatrix{ 0 \ar[r] & \mathop{\mathrm{Ker}}(\sigma ) \ar[r] \ar[d]^ a & \mathbf{Z}[\mathcal{F}] \ar[r] \ar[d] & \underline{\mathbf{Z}} \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \mathcal{H} \ar[r] & \mathcal{E} \ar[r] & \underline{\mathbf{Z}} \ar[r] & 0 }$

Here $\mathcal{E}$ is the extension obtained by pushout. From the long exact cohomology sequence associated to the lower short exact sequence we obtain an element $\xi = \xi _\mathcal {F} \in H^1(X, \mathcal{H})$ by applying the boundary operator to $1 \in H^0(X, \underline{\mathbf{Z}})$.

Conversely, given $\xi \in H^1(X, \mathcal{H})$ we can associate to $\xi$ a torsor as follows. Choose an embedding $\mathcal{H} \to \mathcal{I}$ of $\mathcal{H}$ into an injective abelian sheaf $\mathcal{I}$. We set $\mathcal{Q} = \mathcal{I}/\mathcal{H}$ so that we have a short exact sequence

$\xymatrix{ 0 \ar[r] & \mathcal{H} \ar[r] & \mathcal{I} \ar[r] & \mathcal{Q} \ar[r] & 0 }$

The element $\xi$ is the image of a global section $q \in H^0(X, \mathcal{Q})$ because $H^1(X, \mathcal{I}) = 0$ (see Derived Categories, Lemma 13.20.4). Let $\mathcal{F} \subset \mathcal{I}$ be the subsheaf (of sets) of sections that map to $q$ in the sheaf $\mathcal{Q}$. It is easy to verify that $\mathcal{F}$ is a torsor.

We omit the verification that the two constructions given above are mutually inverse. $\square$

Comment #7114 by Sándor on

You could avoid using the pushout by by applying the boundary operator of the first ses to 1∈H^0(X,Z), which gives an element in H^1(X, Ker(\sigma)), which maps naturally to H^1(X,H). This way the claim that the two operations are inverses to each other seem more straightforward: the Z in the first sequence is the subsheaf of Q generated by q and Z[F] is the subsheaf of I generated by F.

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