Using computations with the hyperalgebra I seem to get that the exact sequence splits precisely when $r+1$ is not divisible by $p$. In fact I am just guessing we are talking about the same exact sequence. Curiously the text in the answer no longer matches the question. Who caused that?

Do you really just want a surjective map, or do you want a split surjective map?

Good filtration theory tells us that there is always a unique submodule of $E \otimes S^r (E)$ that is isomorphic to $S^{r-1} (E)$. But if the characteristic is two and $r$ equals one, the surjection has this submodule in the kernel.

Maybe it helps to observe that the commutator subgroup of $G$ is the same as the commutator subgroup of $\mathbb{R}^{*}G$. And the latter commutator subgroup is $SL_n(\mathbf{R})$.

I should have said that Q1 is wrong for $n=2$.

For $n=2$ it is wrong. Take $T$ of determinant one.

No, because you only give one definition. What you could do is give a definition of $G$-equivariant sheaf that matches.

An example that is not a representation of the algebraic group sends any matrix to its entry wise $p$-th root and views the resulting matrix as describing the action on an $n$-dimensional vector space.

This seems to be about representations of an algebraic group, not the discrete group $\mathrm{SL}(n,\bar{F}_p)$.

@S.A.K.A. John You have changed the wording again. It still does not make sense. Now $G$ has nothing to do with $R$. Is $G$ an abstract group or not? What does "over some ring" mean? My example shows that such things matter. As a real Lie group my example has only one nontrivial normal subgroup, but it should never be written $\mathbb{G}_a$.

Why Smith normal form? Hermite normal form suffices.

Play with the Petersen graph. Its edges have no three coloring. Now `untwist' it.