A famous result of Sullivan (closely related to work of Wilkerson) says that the group of isotopy classes of diffeomorphisms of a simply-connected closed smooth manifold of dimension $\geq 5$ is commensurable with an arithmetic group. Sullivan gave an outline of its proof, and several later papers filled in more details treating some additional cases. These in particular give more of a hint about the required results from the theory of algebraic groups. 


One such paper, Triantafillou's [The Arithmeticity of Groups of Automorphisms of Spaces](http://www.ams.org/books/conm/231/3367/conm231-3367.pdf) contains the following statement regarding a useful general result. It concerns maps from a short exact sequence of groups
$$1 \to A \to A' \to A'' \to 1$$
to a short exact sequence of groups obtained by taking $\mathbb{Q}$-points of an extension of algebraic groups
$$1 \to G \to G' \to G'' \to 1.$$



> 
> To complete the proof one needs an argument to the effect that
>  the middle vertical map of a diagram of short exact sequences involving the $\mathbb{Q}$-points of algebraic groups is arithmetic if the outer maps
>  are arithmetic and the kernel of the algebraic extension is unipotent.
>  Here a map from a group into an algebraic group over $\mathbb{Q}$ is called
>  *arithmetic* if its kernel is finite and its image is arithmetic. It appears
>  that no such result exists in print nor is it folklore in the subject.
>  Upon our inquiry, A. Borel furnished a proof of this algebraic fact
>  which involves basic structure theorems of algebraic groups and the
>  behaviour of lattices under the maps $\exp$ and $\log$ between the Lie
>  algebra and the unipotent group.
> 
> 
> 


However, she does not provide Borel's proof and instead uses a weaker result sufficient for her purposes. Is there a proof of this statement available in the literature somewhere?



A result of Borel on extensions of arithmetic groups
A proof appears in Section 2.3 of my paper [Mapping class groups of manifolds with boundary are of finite type](https://arxiv.org/abs/2204.01945). As pointed out in the comments, it is crucial that $G$ is unipotent.


