You're right, sorry I was rushing. That doesn't show the surjectivity. It's probably not easier to argue on the level of the affine algebra in this case. In any case, I don't really see your problem now. Either you convince yourself that the element you construct in the inverse map depends polynomially on the entries of the coset representative (which to me seems OK). Or you simply use the proof given in 13.22 in DM.

which implies that the induced algebra homomorphism $f^* : \mathbb{A}[\mathcal{L}_{G_1}^{-1}(U_1)/T_1^F] \to \mathbb{A}[\mathcal{L}_G^{-1}(U)/T^F]$ is injective (because $f$ is a dominant morphism). However $\mathbb{A}[\mathcal{L}_G^{-1}(U)/T^F]$ is generated by the indictor maps of the cosets, i.e. $\phi(y) = 1$ if $y \in xT^F$ and 0 otherwise (for a fixed coset $xT^F$). Clearly each such map is the image of a corresponding map under $f^*$. Thus $f^*$ is an isomorphism, so $f$ is an isomorphism.

OK, if you're not happy that the map is an isomorphism of varieties then you can argue as follows (I'll use the notation of DM). As $\mathcal{L}_G(U)/T^F$ is an affine quotient we have the affine algebra of this variety is simply all the regular functions $\phi \in \mathbb{A}[\mathcal{L}_G(U)]$ which are constant on the $T^F$-cosets. Now we've already shown that $f : \mathcal{L}_G^{-1}(U)/T^F \to \mathcal{L}_{G_1}^{-1}(U_1)/T_1^F$ is bijective ...

Also the proof of 13.20 looks fine to me. I think the point is that they're implicitly using Proposition 10.10.

Well, the thing that is easier in the $\theta=1$ case is that you can just look at the cohomology of the quotient variety $\mathcal{L}^{-1}(U)/T^F$. This will be the same as taking the cohomology of $\mathcal{L}^{-1}(U)$ and tensoring with the trivial character. However, for arbitrary $\theta$ one cannot simply replace the tensor product with the cohomology of some variety. This is what makes it more complicated.

What you're trying to show is Proposition 13.22 in Digne-Michel. The statement is proved for Deligne-Lusztig induction from any Levi subgroup. In the statement they (accidentally) assume that the parabolic is F-stable but this assumption is not used in the proof.

The following paper by Aschbacher is probably the canonical reference: Aschbacher, Michael. "On the maximal subgroups of the finite classical groups." Inventiones mathematicae 76.3 (1984): 469-514. There's also a book by Kleidman and Liebeck which gives more details. In fact, §3.5 of that book gives a list of all the subgroups as far as I can tell.

@Jim: I guess the isomorphism makes it implicitly difficult to distinguish them. However, it is also surprisingly useful to use the equivalence of both constructions in the case of finite groups. Maybe the definition by tensor products was preferable because of the universal property of the tensor product, or was just an artefact of who first defined induced representations (Frobenius?).

Do you know if there is some catch theorem classifying when the functors are isomorphic, or fail to be isomorphic, for infinite groups? Or do you just have to examine each situation on its own?

The finite groups case is discussed in detail at this mathstackexchange post math.stackexchange.com/questions/225730/… .

This is just the difference between induction and coinduction right? In particular, whether you think of induction as a right or left adjoint to restriction. These are isomorphic functors for finite groups but for infinite groups I thought these could be different? Maybe this is relevant math.columbia.edu/~woit/LieGroups-2012/inducedreps.pdf .

That's a nice example for $C_p$, I'd never thought about that before.

See for instance (27.21) on pg. 186 of Curtis and Reiner's fantastic book "Representation theory of finite groups and associative algebras".

Character theory provides a very elegant tool when studying representations of finite groups but many statements can be proved without using characters. For instance, the identity you mention follows from the fact that the group algebra $\mathbb{K}G$ is semisimple when $\mathbb{K}$ is an algebraically closed field whose characteristic doesn't divide $|G|$. In particular $\mathbb{K}G$ is isomorphic to a direct sum of matrix algebras $M_{d_i}(\mathbb{K})$. Counting dimensions as $\mathbb{K}$-vector spaces gives you the numerical identity. This can be proved without character theory.

@JimHumphreys Exceptional doesn't necessarily matter but then one can't use the adjoint representation in the other types. For the classical cases, not of type $\mathrm{A}$, one could play the same game with the natural representation. It is true that there are numerous Springer isomorphism but I am looking for just one with good properties. Thus I need something I can get my hands on to show that these good properties hold.