Let $X$ be a finite-type scheme over a field $k$. Let $G$ be a finite-type group scheme over $k$; we write $G\_X$ for the base-change of $G$ from $\operatorname{Spec}(k)$ to $X$.


Suppose $f : Y \rightarrow X$ is a $G\_X$-torsor for the fppf topology (i.e. we have an $X$-group scheme action of $G\_X$ on $Y$ such that the morphism $G\_X \times\_X Y \rightarrow Y \times\_X Y$ given on points by $(g,y) \mapsto (y,gy)$ is an isomorphism). Such a $Y$ gives a class $[Y]$ in the fppf cohomology set $H^1(X,G\_X)$ that classifies fppf $G\_X$-torsor sheaves over $X$ (this is defined with Cech cohomology).


Consider the specialization map
$$
s : X(k) \rightarrow H^1(k,G)
$$
that sends $x : \operatorname{Spec}(k) \rightarrow X$ to the pull-back of $[Y]$ by $x$, which is an element of $H^1(k,G)$. Note: if $G$ is smooth, the fppf cohomology set $H^1(k,G)$ may be identified with the Galois cohomology set $H^1(k,G(k^{\mathrm{sep}}))$.


Supposing it exists, fix a $y \in Y(k)$. We obtain an exact sequence of pointed sets
$$
0 \rightarrow G(k) \rightarrow Y(k) \stackrel{f}{\rightarrow} X(k) \stackrel{s}{\rightarrow} H^1(k,G)
$$
where $G(k) \rightarrow Y(k)$ is just the inclusion of the fiber above $y$. From the looks of it, I'd say that this has to be the start of a long exact sequence of some kind. The question only is: what kind? I don't see an obvious way of continuing it, since $Y$ doesn't necessarily carry any group structure so as to give meaning to the expression $H^1(k,Y)$.


**Question:** Is this exact sequence part of a long exact sequence? For instance, are we witnessing some instantiation of homotopy theory? If not, is there any other more conceptual way of viewing the above sequence?


**Example:** As a motivating example, let me show you why the image of $s$ - and therefore a continuation of the exact sequence from above - is an interesting object of study. Let $E$ be an elliptic curve over $\mathbf{Q}$ given by $y^2=f(x)$. Let $Y$ be $E - E[2]$, let $X$ be $\operatorname{Spec}(\mathbf{Q}[x,f^{-1}])$ (i.e., the affine line with coordinate $x$ and with the subscheme $f=0$ deleted), and let $f:Y \rightarrow X$ be the map that sends $(x,y)$ to $x$. Let $G = \mu\_2$. We endow $Y$ with the structure of a $G$-torsor by letting the non-trivial element of $G$ send $(x,y)$ to $(x,-y)$. Then the image of the map
\begin{align\*}
X(\mathbf{Q}) & \rightarrow \mathbf{Q}^{\ast}/\mathbf{Q}^{\ast 2} ~~ (\cong H^1(\mathbf{Q},\mu\_2)) \\\
x & \mapsto f(x) \pmod{\mathbf{Q}^{\ast}/\mathbf{Q}^{\ast 2}}
\end{align\*}
consists precisely of those elements $c \in \mathbf{Q}^{\ast}/\mathbf{Q}^{\ast 2}$ such that $cy^2=f(x)$ contains rational points other than the "trivial ones", i.e. the zeros of $f$ and the point at infinity.



Do torsors give a long exact sequence of cohomology?
It actually more like the ending of a long exact sequence, rather than the beginning. To see what's going on consider the analogous case in topology. For this you replace the Galois group of $k$ with a discrete group $\Gamma$ and the category of $k$-schemes with the category of $\Gamma$-spaces. Instead of an algebraic group you now have a topological group G equipped with an action of $\Gamma$ on its classifying space BG. A G-torsor is a principle fibration $Y \to X$, which in homotopy theory corresponds to a $\Gamma$-equivariant fibration sequence of the form
$$ Y \to X \to BG $$
Taking $\Gamma$-homotopy fixed points one obtains a fibration sequence
$$ Y^{h\Gamma} \to X^{h\Gamma} \to BG^{h\Gamma} $$
which leads to a long exact sequence of homotopy groups ending with
$$ ... \to \pi\_1(BG^{h\Gamma}) \to \pi\_0(Y^{h\Gamma}) \to \pi\_0(X^{h\Gamma}) \to \pi\_0(BG^{h\Gamma}) $$ 
where $\pi\_1(BG^{h\Gamma}) = \pi\_0(G^{h\Gamma})$. This tail is the homotopy theoretic analogue of the sequence
$$ G(k) \to Y(k) \to X(k) \to H^1(k,G) $$
and so one should consider this sequence as the ending, and not the beginning of a long exact sequence. However, as apposed to the homotopy theoretic analogue, the map $G(k) \to Y(k)$ is always injective, making it seems like the sequence is just starting.


The analogy with the homotopy theoretic case can be made more precise by considering the etale homotopy type (see <http://arxiv.org/abs/1110.0164>).


