Exact sequence of Weyl groups If we note $A_{k}$ the category of affine algebraic groups defined over $k$ and $\mathcal{G}$ the category of finite groups, we have a functor $W:A_{k}\longrightarrow \mathcal{G}$, where $W(G)$ is the Weyl group of an algebraic group $G$. Is that functor exact? 
 A: Yes, it is exact on the category of connected reductive groups and normal homomorphisms over an algebraically closed field $k$.
For an algebraically closed field $k$ and a connected reductive $k$-group $G$, one can construct a canonical based root datum ${\rm BRD}\ G$, so we obtain a canonically defined Weyl group $W(G)=W({\rm BRD}\ G)$. 
Concerning based root data, see Sections 1 and 2 in T.A  Springer, Reductive groups, in: "Automorphic forms, representations and L-functions", Proc. Sympos. Pure Math. 33, part 1, pp. 3-27, Providence 1979. See also Section 1.3 in Brian Conrad, Reductive group schemes, in: "Autour des schémas en groupes", Vol. I, pp. 93–444, Panor. Synthèses, 42/43, Soc. Math. France, Paris, 2014.
I would say  that $G\mapsto W(G)$ is not a functor on the category of connected reductive $k$-groups and homomorphisms of $k$-groups (how can one define  the induced homomorphism of Weyl groups?). However, it is certainly a functor on the category of connected reductive $k$-groups and normal homomorphisms of $k$-groups. A homomorphism of connected reductive $k$-groups is called normal if its image is a normal subgroup. To a normal homomorphism $\phi\colon G_1\to G_2$ one can associate an induced morphism of based root data
$$ \phi_{\rm BRD}\colon\, {\rm BRD}\ G_1\to{\rm BRD}\ G_2$$
and an induced homomorphism of Weyl groups
$$\phi_W\colon\, W({\rm BRD}\ G_1) \to  W({\rm BRD}\ G_2).$$
If we have a short exact sequence of connected reductive $k$-groups
     $$1\to G_1\to G_2\to G_3\to 1,$$
then we obtain an induced short exact sequence of semisimple $k$-groups of adjoint type, which clearly splits, so we obtain a split short exact sequence of Weyl groups. In other words, we obtain a short exact sequence 
  $$1\to W(G_1)\to W(G_2)\to W(G_3)\to 1,$$
where  $W(G_2)=W(G_1)\times W(G_3)$.
