# Exponential map generates the identity component of closed linear groups

In page 15 of "Lie groups beyond an introduction" by Prof. Knapp, Corollary 0.20 states that the exponential map (of linear Lie algebra $$\mathfrak g$$) generates the identity component $$G_0$$ of a closed linear group $$G$$. The proof goes like

1 by continuity of the exponential map $$exp(\mathfrak g)$$ is connected

2 $$exp(\mathfrak g)$$ contains a neighborhood of 1 in the group (local diffeomorphism)

3 the smallest subgroup containing a nonempty open set in $$G_0$$ must be $$G_0$$

3 doesn't look obvious to me though, I don't know how to prove that $$exp(\mathfrak g)$$ is a subgroup. In addition, if we can show it's open, then it's an open closed connected subgroup, so it must be $$G_0$$ since $$G_0$$ is connected.

• Since $\exp \mathfrak{g}$ contains a neighborhood of $1$, then for any $g$ generated by $\mathfrak{g}$, $g \cdot \mathfrak{g}$ contains a neighborhood of $g$ whose elements are also generated by $\mathfrak{g}$. – Travis Willse Apr 8 at 3:50

The image of the exponential map isn't a subgroup in general, it only generates the identity component. See here: (non?)-surjectivity of the exponential map to $SL(2,\mathbb{C})$
• Thank you for your answer, but I am still confused. I think $SL(2,\mathbb{C})$ is simply connected, hence its identity component is the whole group. If there is one element, as shown in the discussion there, that cannot be obtained by exponential map, how can we say the exponential map generates the identity component? – liyiontheway Apr 8 at 5:00
• Well, perhaps I misunderstood "generates". $exp(\mathfrak g)$ is a subset of the identity component, not necessarily a subgroup, but the subset generates the identity component, that is any element in the identity component can be obtained by multiplying finite number of elements in $exp(\mathfrak g)$. Am I correct? – liyiontheway Apr 8 at 5:07