Weyl group of a quotient by a central subtorus. This is probably an easy question, hanging on some minor detail, but i cannot find a proof for it. I try working through T.A. Springer's book "Linear algebraic groups" and i got stuck at remark 7.1.4. (I know there is already a question asked about the same remark, but concerning a different statement made in it). Specifically i am interested in proofing the following:  
Let $G$ be a connected linear algebraic group, $T\subset G$ a maximal subtorus and $S \subset C(G)$ another torus, lying in the center of the group $G$. Then it holds, that:
$$
W(G,T) \cong W(G/S,T/S).
$$
So what i do is:
For $n\in N_G(T)$, one already has $nTn^{-1}=T$, hence also module $S$, which gives a function
$$
\phi : N_G(T)  \to \frac{N_{G/S}(T/S)}{Z_{G/S}(T/S)} \\
n \mapsto n ~\mathrm{mod}~S~,
$$
which is surjective, because $S\subset T$ (i could do that).  
What i need would be a hint on how to proof, that $\ker(\phi) = Z_G(T)$ holds. Thanks in advance!
 A: Check that $N_{G/S}(T/S) = N_G(T)/S$ and $Z_{G/S}(T/S) = Z_G(T)/S$ and use the third isomorphism theorem.
I'll write out the details of why $Z_{G/S}(T/S) = Z_G(T)/S$.  The inclusion '$\supseteq$' is clear.  Conversely, let $gS \in G/S$, and suppose $gS$ commutes with every element of $T/S$.  This means that for all $t \in T$, $gtg^{-1} \in S$.  Then $gtg^{-1}$ commutes with everything in $G$, so 
$$gt = (gtg^{-1})g = g(gtg^{-1})$$
which implies $t = gtg^{-1}$.  Thus $g \in Z_G(T)$, and $gS \in Z_G(T)/S$.
A: $\require{AMScd}$
So i found an answer to this question in Humphrey's "Linear algebraic groups" (around paragraph §25). There a more general proposition is shown, namely:
For an epimorphism $\varphi : G\to G'$, such that $\varphi(T) =: T'$ is the maximal torus of $G'$, and that $\ker(\varphi)$ is contained in every Borel group of $G$, $\mathfrak{B} \to \mathfrak{B}'$ (the induced map on Borel groups), $\mathfrak{B}^T \to \mathfrak{B}'^{T'}$ (induced map on Borel groups over $T$) and $W(G,T) \to W(G',T')$ (induced map on Weyl groups (like $\phi$ in the question)) are bijective.
The point being, that $W(G,T) \cong \mathfrak{B}^T$ and that there is a commutative diagram
$$
\begin{CD}
W(G,T) @>>> W(G',T') \\
@VV{\cong}V @VV{\cong}V\\
\mathfrak{B}^T  @>>> \mathfrak{B}'^{T'},
\end{CD}
$$
where the horizontal arrows are induced by the epimorphism $\varphi$.
I will show bijectivity of the lower horizontal arrow:
It suffices to show injectivity for $\mathfrak{B} \to \mathfrak{B}'$. So take two Borel groups of $G$, call them $B_1$ and $B_2$, s.t. $\varphi(B_1)=\varphi(B_2)$. Then by the conjugacy theorem $B_1= gB_2g^{-1}$, for some $g\in G$. Then it holds that $\varphi(B_1) =\varphi(g)\varphi(B_1)\varphi(g)^{-1}$, i.e. $\varphi(g)\in N_{G'}(\varphi(B_1)) =\varphi(B_1)$. Since $\ker(\varphi) \subseteq B_1$, $g\in B_1$ and hence $B_1=B_2$ follows.
For surjectivity take $B'$ Borel group of $G'$, over $T'$. Then $\varphi^{-1}(B')$ is some subgroup $H$, s.t. $G/H \to G'/B'$ is an epimorphism. Since $G'/B'$ is complete (Borel group is parabolic) $G/H$ is complete, i.e. $H$ is a parabolic and contains at least one Borel group $B$. Now $\varphi(B) \subseteq B'$ is a Borel group inside  a bigger one, hence by the conjugacy theorem, they agree.
This shows that $W(G,T)\to W(G',T')$ is bijective.  
Now to show that the conditions of Humphrey are actually met, by the ones given in the question:  


*

*$S\subseteq C(G)^0 \subseteq C(B) \subseteq B$, i.e. the identity component of the center of $G$ lies in the center of every Borel group $B$.

*$S$ lies in $T$, because $S$ is in $Z_G(T)^0$ (it commutes with every element of $T$) and thus there is a maximal torus in the Cartan subgroup of $T$ over $S$. Since there is only one maximal torus in $Z_G(T)^0$ and it is $T$, $S\subseteq T$.  


I will omit the rest, as it is again contained in Springer's book.
