Do torsors give a long exact sequence of cohomology?

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.

-
If $G \to Y \to X$ is an exact sequence of group schemes, then $Y \to X$ is a $G$-torsor, and your exact sequence is part of the standard long exact sequence of group cohomology. So an answer would presumably be a generalization of that exact sequence to other cases. I do not know anything about the homotopy-theory or other considerations that would enable someone to generalize this to higher degree cohomology groups. – Will Sawin Feb 4 '13 at 20:20
Thank you, Will, this is indeed part of the motivation for my question. I should have included it in the post. – René Feb 4 '13 at 23:21

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.