Identity component and rational point over algebraically closed field For an algebraic group $G$ over $\mathbb{C}$ (in general, an algebraically closed field with character $0$), let the identity component of G be denoted by $G_{0}$. I wonder whether the rational point of $G_0$ equals the identity component of $G$'s rational point, namely $G_0(\mathbb{C})=G(\mathbb{C})_0$, holds.
Here, the topology is defined with regard to the fact that rational point of an algebraic group over $\mathbb{C}$ forms an Lie group over $\mathbb{C}$.
 A: This is true if, as reuns suggests, what you mean is the following:

Fact: Let $G$ be an algebraic group over $\mathbb{C}$. Then, $(G^\circ)^\mathrm{an}=(G^\mathrm{an})^\circ$.

Here I am using $\circ$ to denote the connected component (of an algebraic group or a topological group) I am using $X^\text{an}$ to denote the analytification of a variety $X$ over $\mathbb{C}$ (i.e. equipping $X(\mathbb{C})$ with the usual structure of a complex analytic space).
The reason for this is simple. To show that $(G^\circ)^\text{an}$ is the connected component of $G^\text{an}$ we need to show that"

*

*$(G^\circ)^\text{an}$ is connected.

*$G^\text{an}/(G^\circ)^\text{an}$ is finite.

The first of these is a general property of the analytification functor in general (e.g. see this post) and for the second we note that $G/G^\circ$ is finite and eviently we have the following equality of sets
$$G^\text{an}/(G^\circ)^\text{an}=G(\mathbb{C})/G^\circ(\mathbb{C})=(G/G^\circ)(\mathbb{C})$$
from where the conclusion follows.
