TITLE: Differential of the inversion of Lie group
QUESTION [5 upvotes]: Let $G$ be a Lie group and $\iota \colon G\to G$ denote the inversion. If $e$ is the identity of $G$, prove that: $$\text d \iota _e = -\text {id} _{T_e G}.$$
I understand that the differential at $e$ should be an involution, since $$\iota \circ \iota = \text{id}_G$$
implies that $$\text d \iota _e \circ \text d \iota _e=\text d (\iota \circ \iota)_e=\text d(\text {id} _G)_e=\text {id} _{T_e G}.$$
However I don't know how to conclude the thesis from this. Any hint?

REPLY [4 votes]: It can be sooo simple... :)
Applying the multiplication infinitesimally:
$$[\alpha]+\mathrm{d}\iota[\alpha]=[\alpha]+[\iota\circ\alpha]=\mathrm{d}\mu[(\alpha,\alpha^{-1})]=[\mu(\alpha,\alpha^{-1})]=[e]=0$$
(Note that all equality signs in this line are for real.)