Given an action of a group $G$ on a topological space $X$, the associated *homotopy quotient* is $$X_G := (EG \times X)/G,$$ where $EG$ is the total space of a universal principal $G$-bundle $EG \to BG$ and the quotient is by the (free) diagonal action of $G$ on $EG \times X$.
The $G$-*equivariant cohomology* of $X$ is defined to be $$H^\ast_G(X) := H^\ast(X_G),$$ the cohomology of $X_G$ (say singular with $\mathbb{Q}$ coefficients).

The homotopy quotient $X_G$ is the total space of a $X$-bundle over $BG$, given by $[(e,x)] \mapsto eG$, so the fiber inclusion $X \to X_G$ induces a restriction homomorphism $$H_G^\ast(X) \to H^\ast(X).$$ One says the $G$-space $X$ is *equivariantly formal* if this homomorphism is surjective.

A compact Lie group $G$ has a maximal torus $T$, which acts on $G$ by conjugation. I would like to find a class in $H^{\dim G}_T(G)$ that restricts to a generator for the top-dimensional cohomology $H^{\dim G}(G)$. Is that always possible? More generally, I am curious:

Is $G$ an equivariantly formal $T$-space?

So far, I only know this for the abelian case $G = T$, where $G_T = T_T = (ET \times T)/T = BT \times T$ and a Künneth formula applies.

**Edit**:

I see from this question that $H^\ast_G(G) = H^\ast(G) \otimes H^\ast(BG)$, but don't understand the proof there well enough to see if or how it could carry over. Anyway, thanks.