Relations between center (fundamental group) and (co)root and weight lattices for Lie groups

I would like to find some explanation or reference for the following facts, provided they are correct, and clarify some of the assumptions. Denote by $G$ a (perhaps semisimple compact connected) Lie group:

1. its center is given by the quotient of the weight lattice by the root lattice, i.e. $Z(G)=\Lambda_\text{weight} / \Lambda_\text{root}$

   1a. if moreover $G$ is (perhaps simple and simply connected) of ADE type, its center is $Z(G)=\Lambda^*_\text{root} /\Lambda_\text{root}$, where $*$ denotes the dual lattice

2. its first homotopy group is given by the quotient of the co-weight lattice by the co-root lattice, i.e. $\pi_1(G)=\Lambda_\text{weight}^\vee / \Lambda_\text{root}^\vee$

To see such relations more clearly, one should first of all avoid identifying the weight space $\Lambda\otimes\Bbb R$ with its dual space. Even in the semisimple case, an invariant inner product is only determined up to a scalar factor, and although the Killing form provides a distinguished choice, it is really not that special, nor well behaved under direct sum decompositions.

Let $G$ be a compact connected semisimple Lie group with Cartan subgroup $H$, let $\def\Hom{\operatorname{Hom}}\Lambda=\Hom(H,\Bbb C^\times)$ be its character lattice. With $\def\h{\mathfrak h}\h$ the Lie algebra of $H$, each $\lambda\in\Lambda$ has a derivative $D\lambda:\h\to\Bbb C$ with purely imaginary image, and by mapping $\lambda$ to $\def\1{2\pi\mathbf i}{D\lambda\over\1}:\h\to\Bbb R$, we can embed $\Lambda$ into to dual vector space $\h^*$. For all $h\in\h$ one has $\lambda(\exp(h))=\exp(\1\langle\lambda,h\rangle)$ (the pairing indicates that the second $\lambda$ is being interpreted as linear form on $\h$). Therefore $\ker(\exp:\h\to H)$ is the set of $h\in\h$ on which all linear forms in $\Lambda\subset\h^*$ take integral values; this is a lattice in$~\h$ that I shall denote by $\Lambda^\vee$. We have $H\cong\h/\Lambda^\vee$.

The conjugation action of $G$ on itself induces (as its derivative at the identity) the adjoint action $\def\g{\mathfrak g}G\to GL(\g)$, a linear representation of $G$ on its Lie algebra$~\g$. By connectedness of$~G$, an element $g\in G$ is central whenever its adjoint action is trivial. The centre $\def\ZG{\mathrm Z(G)}\ZG$ is contained in $H$. Restricting the adjoint action to$~H$, the space $\g$ decomposes into the direct sum of $\h$ (the set of fixed points, or weight space for the zero weight) and of the one-dimensional root-subspaces. The characters associated to the root subspaces are the roots, which form a finite subset$~\Phi$ of$~\Lambda$. It follows that $\ZG=\bigcap_{\alpha\in\Phi}\ker(\alpha)$ (where the kernel is of $\alpha$ as element of $\Hom(H,\Bbb C^\times)$).

The $\Bbb Z$-span $\langle\Phi\rangle$ of $\Phi$ is a sub-lattice of$~\Lambda$ that has full rank, and therefore finite index (only at this point is it relevant that $G$ is semisimple; until now the weaker hypothesis of reductive suffices). It is already spanned by a chosen subset of simple roots. In the correspondence $H\cong\h/\Lambda^\vee$, a subset $\ker(\alpha)\subset H$ corresponds to the set of classes of $h\in\h$ with $\langle\alpha,h\rangle\in\Bbb Z$. Therefore $\ZG$ corresponds naturally to $\langle\Phi\rangle^\vee/\Lambda^\vee$, where $\langle\Phi\rangle^\vee$ is the dual lattice to the root lattice$~\langle\Phi\rangle^\vee$, the set of $h\in\h$ for which $\langle\alpha,h\rangle\in\Bbb Z$ for all simple roots$~\alpha$, and therefore for all $\alpha\in\langle\Phi\rangle$. This dual lattice is a sub-lattice of$~\h$ that contains $\Lambda^\vee$ as a finite index sub-lattice, so $\ZG$ is a finite Abelian group.

To get back to the original question, one sees that the quotient $\Lambda/\langle\Phi\rangle$ is naturally isomorphic not to $\ZG$, but to its dual group $\widehat{\ZG}=\Hom(\ZG,\Bbb C^\times)$. Every finite Abelian group is isomorphic to its dual group (by the structure theorem), but not naturally so. So the first statement is not wrong, it is just not the best way to represent $\ZG$ as a quotient of lattices.

For the fundamental group $\pi_1(G)$ one needs to consider the universal cover $\widetilde G$ of $G$, which is a simply connected group. As usual for universal covers, the fundamental group $\pi_1(G)=\pi_1(G,e)$ is isomorphic the fibre of the covering $\widetilde G\to G$ over the identity $e$ of$~G$ (by associating to a loop in$~G$ the endpoint of its lift to$~\widetilde G$). Since the inverse image of $\ZG$ is $\mathrm Z(\widetilde G)$, the latter centre contains the mentioned fibre (which is the inverse image of $\{e\}$). So $\pi_1(G,e)$ can be seen a subgroup of $\mathrm Z(\widetilde G)$, namely the kernel of the natural surjection $\mathrm Z(\widetilde G)\to\ZG$. But by the above $\mathrm Z(\widetilde G)$ is isomorphic to the quotient of lattices $\langle\Phi\rangle^\vee/\widetilde\Lambda{}^\vee$, where $\widetilde\Lambda{}^\vee$ is the dual lattice to the character lattice $\widetilde\Lambda$ of$~\widetilde G$, the lattice generated by the fundamental weights. But we have $\widetilde\Lambda{}^\vee=\langle\Phi^\vee\rangle$, the lattice generated by the (simple) coroots $\alpha^\vee$ (and not to be confused with the dual lattice $\langle\Phi\rangle^\vee$ of the root lattice). Then $\pi_1(G)$ gets naturally identified with the kernel of the projection $\langle\Phi\rangle^\vee/\langle\Phi^\vee\rangle\to\langle\Phi\rangle^\vee/\Lambda{}^\vee$, which is $\Lambda{}^\vee/\langle\Phi^\vee\rangle$. This gives you your second point, but you should not write the coroot lattice as $\Lambda_{\rm root}^\vee$: this would suggest the larger lattice $\langle\Phi\rangle^\vee$ rather than $\langle\Phi^\vee\rangle$.

Your point 1a. seems just confused to me; the only way it could be justified is by somehow making $\Lambda_{\rm root}^\vee$ denote $\Lambda$, but any attempt to do so (under rather strong hypothesis) seems pointless, as $\Lambda$ is quite clear.