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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 taht functor is exact ?

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The formulation is incomplete: Is the field arbitrary, and how does it affect the question if it isn't algebraically closed? Are your affine algebraic groups required to be connected, or not (and does it matter)? In any case, what precisely do you mean by "the Weyl group" of the given group? And by "exact" do you mean left or right or both? The question needs more context as well. (Why is it interesting one way or the other?) –  Jim Humphreys Apr 1 '12 at 22:24
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up vote 3 down vote accepted

Yes, it is exact.

For an algebraically closed field $k$ and a connected reductive $k$-group $G$, one can construct a canonical based root datum of $G$, so we obtain a canonically defined Weyl group $W(G)$. I am not sure that $G\mapsto W(G)$ is a functor on the category of connected reductive $k$-groups and homomorphisms of $k$-groups. However, it is certainly a functor on the category of connected reductive $k$-groups and normal homomorphisms. A homomorphism of connected reductive $k$-groups is called normal if its image is a normal subgroup.

If we have a short exact sequence of connected reductive $k$-groups, then we obtain an induced short exact sequence of semisimple groups of adjoint type, which clearly splits, so we obtain a split short exact sequence of Weyl groups. In other words, if we have a short exact sequence $1\to G_1\to G_2\to G_3\to 1$ of connected reductive $k$-groups, then $W(G_2)=W(G_1)\times W(G_3)$.

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