Let $G$ be a Lie group, $G_0$ the connected component of the identity, $\pi:G\to G/G_0$ the quotient map onto the component group, and $C_G(G_0)$ the centralizer of $G_0$ in $G$.
Question: Is it always the case that $\pi(C_G(G_0))=G/G_0$? If not, is there a "nice" non-trivial (e.g., not just the connected ones) class of Lie groups $G$ always having this property?
I suspect this is false in general, but there seem to be many natural examples where this occurs, such as in $\mathrm{GL}(n,\mathbb{R})$, which has the stronger property that $Z(G)$ maps onto $G/G_0$, or in the product of a connected Lie group with a discrete group.
No, the centralizer need not map surjectively onto the component group.
One family of counterexamples is proved by the Lie groups $O(2n) = \{A\in M_{2n}(\mathbb{R}): AA^t = A^tA = I\}$, which has the identity component $SO(2n) = \{A\in O(2n): \det(A) = 1\}$.
Given any $A\in SO(2n)$ and any eigenvalue $\lambda$ of $A$, $B$ permutes the $\lambda$-eigenspaces of $A$: if $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $$\lambda (Bv) = B\lambda v = BAv = A(Bv).$$
For most $A$, these eigenspaces are $1$-dimensional, which implies that $B$ acts as $\pm 1$ on each one. Doing this for all $A$ easily implies that every vector is an eigenvector for $B$, so $B = \pm I$.
But since $2n$ is even, $\det(\pm I) = 1$, so this means $B\in SO(2n)$. That is, $C_{O(2n)}(SO(2n))\subseteq SO(2n)$.