Relative weight lattice Let $G$ be a reductive group over an algebraically closed field $k$. Let $T$ be a maximal torus, $B$ be a Borel subgroup and $I_G$ is the set of simple roots. Let $P$ be a standard parabolic subgroup, $M$ be its Levi containing $T$ and let $I_M$ be the set of simple roots of $M$ (with the natural choice of Borel of $M$). Let $\Lambda_G$ be the weight lattice of G. We define $\Lambda_{G,P}:=\frac{\Lambda_G}{\text{span of $\alpha_i$, $i\in I_M$}}$.

Is $\Lambda_{G,P}=X(Z(M)^0)$, where $Z(M)^0$ is the component of $Z(M)$ (
  the center of $M$)  containing the identity and $X(Z(M)^0)$ is the character group of the
  torus $Z(M)^0$?

 A: I'll use $\mathrm X^*$ instead of $X$ for character lattices, since I can never remember which is which in the $X$/$Y$ notation.  I have also updated this answer from its original wrong formulation to a hopefully correct one.
$\DeclareMathOperator\srank{srank}$Note that $\Lambda_{G, P}$ is a lattice of rank $\srank(G) - \srank(M)$, where $\srank$ stands for the semisimple rank.
Exactly as written, the answer is 'no'; for example, if $G = M$ is a non-trivial torus, then $\Lambda_{G, P}$ is trivial but $\mathrm X^*(\mathrm Z(M)^\circ) = \mathrm X^*(G)$ is not.
If $G$ is semisimple, then $\srank(G) - \srank(M) = \dim(\mathrm Z(M)^\circ)$, so that $\Lambda_{G, P}$ and $\mathrm X^*(\mathrm Z(M)^\circ)$ are lattices of the same rank, hence abstractly isomorphic.  However, there is a natural map $\Lambda_{G, P} \to \mathrm X^*(\mathrm Z(M)^\circ)$ given by restriction, and it need not be an isomorphism; its image may have finite index.  Consider the case $G = \mathrm{SL}_2$ and $M = T$.
$\DeclareMathOperator\Span{Span}$If $G$ is adjoint, then $\Lambda_G = \mathrm X^*(T)$ and $\Span_{\mathbb Z} \{\alpha : \alpha \in I(M, B \cap M, T)\}$ is the annihilator of $\mathrm Z(M)$ in $\mathrm X^*(T)$, so that the restriction map $\Lambda_{G, P} \to \mathrm X^*(\mathrm Z(M))$ is an isomorphism.  Since $\Lambda_{G, P}$ is torsion free, so is $\mathrm X^*(\mathrm Z(M))$, which means that $\mathrm Z(M)$ is connected, and hence we have finally that $\Lambda_{G, P} \to \mathrm X^*(\mathrm Z(M)) = \mathrm X^*(\mathrm Z(M)^\circ)$ is an isomorphism in this case.
