Suppose $G=k^\times$ where $k$ is an algebraic closed field of characteristic $0$(I'm not sure whether this assumption is necessary). Then show that $(\quad)^G$ is an exact functor i.e. given an exact sequence of $k-$algebras $A\to B\to C$ on which $G$ has an action compatible with the homomorphisms between them, the $G-$invariant components still preserves the exactness i.e. $A^G\to B^G\to C^G$ is exact.
In Teleman's paper The quantization conjecture revisited, he wrote $(\quad)^G$ is exact since $G$ is reductive. I know $k^\times$ is reductive but I have no ideas about the relation to exactness. Another remark is that the category in which $A,B$ and $C$ lies might not have to be $k-$algebra but I believe the spirits are totally the same. I'm not familiar with terminology of Lie groups and manifolds so I hope you can give me some algebraic interpretations as far as you can. Thank you for the help!
I have some more ideas. Is it possible that I define a "projecter" using the Lie structure of $k^\times$, just like the operator $\frac{1}{|G|}\sum\limits_{g\in G}g$? I do not know well about this piece of knowledge.