in Herzig's lecture notes on linear algebraic groups (page 24/25, see https://www.math.toronto.edu/~herzig/lin-alg-groups17-seaton-notes.pdf), there is a quite interesting fact on the exactness of diagonalisable linear algebraic groups:
Namely that a sequence $0 \to G_1 \to G_2 \to G_3 \to 0$ is exact
if (by definition)
- it is set theoretically exact and
- the induced sequence on Lie algebras $0 \to Lie(G_1)\to Lie(G_2) \to Lie(G_3) \to 0$ is exact.
Note that $G_i$ are linear algebraic groups over an algebraically closed field $k$. Furthermore, $X^*(G_i)$ is the group of characters given by Hom$_\text{lin.alg.groups}(G_i,G_{m,k})$ with $G_{m,k}$ the multiplicative group of $k$.
Now it is clear that for diagonalisable groups that the Lie algebras are given by Hom$(X^*(G_i),k)$. However, he now claims the sequence (of Lie algebras, I guess) is exact if and only if
$0 \to X^*(G_3) \to X^*(G_2) \to X^*(G_1) \to 0$ is exact.
I don't see why this should hold; my ideas are the following:
We have that Hom(_,k) is a left-exact contravariant functor, so left exactness should be fine by category theory. However, for the functor to be exact, we need that $X^*(G_1)$ is an injective abelian group.
However, all I can prove is that $X^*(G_1)$ is free and thus projective, however this does not imply that Hom( _ ,k)is exact but rather that Hom(k, _ ) is exact.