Question about Tamafumi Kaneyama's Paper: “On Equivariant Vector Bundles On An Almost Homogeneous Variety”

I have two question about Proposition 3.3.:

Proposition3.3. says that $1 \rightarrow Aut(E) \xrightarrow{j} G(E) \xrightarrow{p} T \rightarrow 1$ is a short exact sequence of algebraic groups over an algebraically closed field $k$, where T is a torus over $k$ and $E$ is a vector bundle of a almost homogeneous variety $X$. I have no question about exactness, but the author says that since both $Aut(E)$ and $T$ are linear algebraic groups, $G(E)$ is a linear algebraic group.

Question1.

Is that means that if $1 \rightarrow G \xrightarrow{f} H \xrightarrow{g} K \rightarrow 1$ is a short exact sequence of algebraic groups, then $H$ is linear if, and only if $G$ and $K$ are linear? Is that easy to see?

In the proof of Proposition3.3., the author says that a surjective algebraic group homomorphism from a torus to a torus always has a section (see Borel [1]).

Question2.

I already found this reference :A. Borel, "Linear algebraic groups", Benjamin (1969) MR0251042 Zbl 0206.49801 Zbl 0186.33201. But I can't find where prove this fact. Who can tell me where is the proof?

Thank you very much!

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To answer your Question 1, note first that if $L$ is a linear group and $A$ is an abelian variety, then all group homomorphisms $A\to L$ and $L\to A$ are constant. The first part is easy because the image of $A$ in $L$ is closed hence affine, and is proper being image of a proper variety, so the image is just one point. For the second part, I don't have an easy proof. One can say either that $L/\mathrm{ker}(L\to A)$ is affine and closed in $A$, hence reduced to one point, or that $L$ is rational, and any morphism from a rational variety to an abelian variety is constant. This can also considered as part of Chevalley's follow theorem: over a perfect field, any smooth connected algebraic group $G$ is uniquely the extension of an abelian variety $A(G)$ by a smooth linear algebraic group $L(G)$: $$1\to L(G)\to G \to A(G) \to 1.$$
The claim in Question 2 (as you stated) is not correct: consider the isogeny $x\to x^2$ in the simplest torus $\mathbb G_m$. It doesn't have a section. In fact, if $T_2\to T_1$ is a surjective group homomorphism of tori, then there exists a section up to isogeny: $T_1\to T_2$ whose composition with $T_2\to T_1$ is given by $x\to x^n$ for some positive integer $n$. This can be proved using the exact sequence $$1\to H\to T_2 \to T_1 \to 1$$ where the connected component of $1$ in $H$, endowed with the reduced structure, is a torus $T_3$. If we denote by $\chi(T)$ the group $\mathrm{Hom}(T, \mathbb G_{m})$ of characters of $T$ (the base field $k$ is algebraically closed here, otherwise, consider the characters over $\bar{k}$), then $$0\to \chi(T_3) \to \chi(T_2)\to \chi(T_1) \to 0$$ is exact up to tensoring by $\mathbb Q$. The exact sequence splits up to tensoring by $\mathbb Q$. By the equivalence of categories between the tori and the finite rank free $\mathbb Z$-modules, $T_2$ is isogeneous to $T_1\times T_3$, so the morphism $T_2\to T_1$ has an "isogeny" section.
Excuse me, I have questions: 1."The exact sequence splits up to tensoring by $\mathbb{Q}$." Dose it means splits by vector spaces over $\mathbb{Q}$? 2.Sorry, could you explain the meaning of "isogeny" section? I can't find proper reference about this concept and definition. Thank you very much! – Peter Hu May 21 '12 at 16:14
@PeterHu: By spliting up to tensoring by $\mathbb Q$, I mean the complex tensored over $\mathbb Z$ by $\mathbb Q$ is an exact sequence (of groups of vector spaces, it doesn't really matter). An isogeny here means a finite surjective homomorphism of (connected) algebraic groups. An "isogeny section" (not really standard terminology) of $f: X\to Y$ means a morphism $g : Y\to X$ such that $g\circ f$ is the multiplication map $Y$ by some positive integer. – user18119 May 22 '12 at 19:51