**Context:**
Let $\pi: \widehat{G} \rightarrow G$ be a surjective morphism between connected reductive groups defined over $\mathbb{F}\_q$ whose kernel is a central torus. Then $\pi : \widehat{G}^F \rightarrow G^F$ is surjective, where $F$ denotes the Frobenius morphisms inducing the $\mathbb{F}\_q$-rational structures.
We thus have an inflation map $- \circ \pi: \text{Irr}(G^F) \rightarrow \text{Irr}(\widehat{G}^F)$. We also have a dual morphism $\pi^\* : G^\* \rightarrow \widehat{G}^\*$ which is injective under the above assumptions. 


**Question:** Is the inflation map compatible with Lusztig series in the sense that for a semisimple $s \in (G^\*)^F$ and any $\chi$ in the Lusztig series $\mathcal{E}(G^F,[s])$ associated to $s$ we have $\chi \circ \pi \in \mathcal{E}(\widehat{G}^F,[\pi^\*(s)])$?


**Motivation:**
It is easy to see that every element of $\mathcal{E}(\widehat{G}^F,[\pi^\*(s)])$ is inflated from $G^F$, so if the answer to my question is positve, this would result in a bijection between $\mathcal{E}(G^F,[s])$ and $\mathcal{E}(\widehat{G}^F,[\pi^\*(s)])$. 


**My Approach:**
It seems natural to me to try to show that inflation is compatible with Lusztig induction, i.e. $(R^G\_T \theta) \circ \pi = R^{\widehat{G}}\_{\pi^{-1}(T)} (\theta \circ \pi)$ for any rational maximal torus $T$ and any $\theta \in \text{Irr}(T^F)$. To show this one could try to compare the Deligne-Lusztig varieties $Y\_U = \{g \in G \:|\: g^{-1}F(g) \in U\}$ and $Y\_{\pi^{-1}(U)}$ for $U$ the unipotent radical of a Borel subgroup of $G$ containing $T$.
This looks promising to me as $\pi$ induces a surjective morphism $Y\_{\pi^{-1}(U)} \rightarrow Y\_{U}$ whose fibres are cosets of $\ker(\pi)^F$. As I see it it would suffice to show that this morphism gives an isomorphism between the $\ker(\pi)^F$-quotient of $Y\_{\pi^{-1}(U)}$ and $Y\_{U}$ but I am unable to prove it.


I tried to mimic the proof of Proposition 13.20 in Representations of Finite Groups of Lie Type by Digne and Michel where the case $s = 1$ is studied for a more general $\pi$.
But unfortunately and contrary to what is claimed there the proof given there does not establish an isomorphism like the above as far as I can tell. Any help is appreciated.



A bijection between Lusztig series induced by inflation
Your approach is correct and is proven in the book by Digne-Michel (in fact a more general statement is proven there). Indeed, by Proposition 13.22 in Digne-Michel we know that


$$R\_{T\subseteq B}^G(\theta)\circ \pi = R\_{\widehat{T}\subseteq \widehat{B}}^{\widehat{G}}(\theta\circ\pi)$$


where $\widehat{T} = \pi^{-1}(T)$ and $\widehat{B} = \pi^{-1}(B)$. Now you just need to show that the conjugacy class of $\theta\circ\pi$ corresponds to the conjugacy class of $\pi^\*(s)$ under duality.


By the definition of duality we have an isomorphism $\lambda : X(T) \to Y(T^\*)$ between the character group of $T$ and the cocharacter group of $T^\* \leqslant G^\*$, which is a dual torus. Furthermore, $\lambda$ is compatible with the Frobenius endomorphism.


According to Proposition 13.7 in Digne-Michel there exists a character $\chi \in X(T)$ of $T$ such that $\theta = \chi|\_{T^F}$, i.e., $\theta$ is the restriction of $\chi$. Now let $\gamma = \lambda(\chi)$ be the cocharacter corresponding to $\chi$. If we assume that $n \geqslant 1$ is large enough so that $T$ is a split torus with respect to the Frobenius endomorphism $F^n$ then we have $s = N\_{F^n/F}(\gamma(\zeta))$ where $\zeta$ is a $(q^n-1)$th root of unity and $N\_{F^n/F} : T^\* \to T^\*$ is the norm map.


Now let us assume that we have chosen an isomorphism $\widehat{\lambda} : X(\widehat{T}) \to Y(\widehat{T}^\*)$ extending $\lambda$. In other words, we have $\widehat{\lambda}(\chi\circ\pi) = \pi^\*\circ\lambda(\chi)$. Now, clearly $\theta\circ\pi = (\chi\circ\pi)|\_{T^F}$ and $\widehat{\lambda}(\chi\circ\pi) = \pi^\*\circ\lambda(\chi) = \pi^\*\circ\gamma$. As $\pi^\*$ clearly commutes with the norm map this gives us that $\theta\circ\pi$ corresponds to $\pi^\*(s) = N\_{F^n/F}(\pi^\*(\gamma(\zeta)))$. Hence


$$R\_{T^\*}^G(s)\circ \pi = R\_{\widehat{T}^\*}^{\widehat{G}}(\pi^\*(s))$$,


which is what you want.


