A question about unipotent characters

Question

I have just started learning (complex) character theory of groups of Lie type and there are some misunderstandings for me in the subject. It is well-known that unipotent characters of $PGL_{n_i}(q_i)$ (and $PSL_{n_i}(q_i)$), where $q_i=p^{r_i}$, are in one-to-one correspondance with the partitions of number $n_i$. My question is about unipotent characters of some specific groups lying between the finite direct products $\coprod_{i} PSL_{n_i}(q_i)$ and $\coprod_{i} PGL_{n_i}(q_i)$, namely the centralizers of semisimple elements.

If we consider a semisimple element $s \in SL_n(q)$, then there exist inegers $n_i$ and $r_i$ with $\Sigma n_ir_i=n$ such that $PSL_{n_1}(q^{r_1})\times...\times PSL_{n_k}(q^{r_k}) \leq \overline{C_{SL_n(q)}(s)} \leq PGL_{n_1}(q^{r_1})\times...\times PGL_{n_k}(q^{r_k})$, where $\overline{C_{SL_n(q)}(s)}$ is the image of $C_{SL_n(q)}(s)$, the centralizer of $s$ in $SL_n(q)$, under the canonical homomorphism $-:\coprod_{i=1}^{k}{GL_{n_i}}\rightarrow \coprod_{i=1}^{k}{PGL_{n_i}}$.


Let $PSL_{n_1}(q_1)\times...\times PSL_{n_k}(q_k) \leq H \leq PGL_{n_1}(q_1)\times...\times PGL_{n_k}(q_k)$, such that $H=\overline{C_{SL_n(q)}(s)}$ for a semisimple element $s\in SL_n(q)$, where $n=\Sigma n_ir_i$ and $q_i=q^{r_i}$. ($-$ is canonical morphism as above) . Also let $\mathbb{U}(G)$ denotes the set of unipotent characters of group $G$.

Question: Is it true that $\mathbb{U}(PSL_{n_1}(q_1)\times...\times PSL_{n_k}(q_k))= \mathbb{U}(H)$?

I know that by a result of George Lusztig, all unipotent characters of $PSL_{n_i}(q_i)$ are extendable to $PGL_{n_i}(q_i)$, and also unipotent characters of $PGL_{n_i}(q_i)$ remain irreducible upon restriction to $PSL_{n_i}(q_i)$. This implies that $\mathbb{U}(PGL_{n_1}(q_1)\times...\times PGL_{n_k}(q_k))=\mathbb{U}(PSL_{n_1}(q_1)\times...\times PSL_{n_k}(q_k)) \subset \mathbb{U}(H)$.
Now assuming that the claim in the question is true, how can I show that $H$ does not have any other unipotent character outside $\mathbb{U}(PSL_{n_1}(q_1)\times...\times PSL_{n_k}(q_k))$? If the claim is false, would you please provide a counterexample?
I would be grateful for any help.