Given a Lie group.
Multiplication and inversion act infinitesimally at the identity by: $$\mathrm{d}\mu:\mathrm{T}_{(e,e)}(G\times G)\to\mathrm{T}_eG:(u,v)\mapsto u+v$$ $$\mathrm{d}\iota:\mathrm{T}_eG\to\mathrm{T}_eG:w\mapsto -w$$
How to prove these statements from scratch?
Ok I got it meanwhile...
Inversion: Geometric Approach
Applying the multiplication infinitesimally gives: $$[\alpha]+\mathrm{d}\iota[\alpha]=[\alpha]+[\iota\circ\alpha]=\mathrm{d}\mu[(\alpha,\alpha^{-1})]=[\mu(\alpha,\alpha^{-1})]=[e]=0$$ (Note that all equality signs in this line are for real.)
Multiplication: Geometric Approach
One has the explicit identification: $$\mathrm{T}_eG\oplus\mathrm{T}_eG\cong\mathrm{T}_{(e,e)}(G\times G):\quad([\alpha],[\beta])\widehat{=}[(\alpha,\beta)]$$
So that tangent curves split accordingly into: $$[(\alpha,\beta)]\widehat{=}([\alpha],[\beta])=([\alpha],0)+(0,[\beta])\widehat{=}[(\alpha,e)]+[(e,\beta)]$$
Exploiting linearity one obtains: $$\mathrm{d}\mu[(\alpha,\beta)]=\mathrm{d}\mu[(\alpha,e)]+\mathrm{d}\mu[(e,\beta)]=[\mu(\alpha,e)]+[\mu(e,\beta)]=[\alpha]+[\beta]$$ (Note that the above identification is rather hard to check.)
Multiplication: Algebraic Approach
Consider the sections: $$s_1:G\to G\times G:p\mapsto (p,e)\quad s_2:G\to G\times G:q\mapsto(e,q)$$
This makes the identification explicit: $$\Psi:\mathrm{T}_eG\oplus\mathrm{T}_eG\to\mathrm{T}_{(e,e)}(G\times G):(\delta,\varepsilon)\mapsto \mathrm{d}_es_1\delta+\mathrm{d}_es_2\varepsilon$$
Do the key observation on sections: $$\mu\circ s_1=\mathrm{id}\quad\mu\circ s_2=\mathrm{id}$$
Applying this on test functions gives: $$\left[\mathrm{d}\mu(\Psi(\delta,\varepsilon))\right]\varphi=\left[\mathrm{d}\mu\mathrm{d}s_1\delta+\mathrm{d}\mu\mathrm{d}s_2\varepsilon\right]\varphi =\delta(\varphi\circ\mu\circ s_1)+\varepsilon(\varphi\circ\mu\circ s_2)=\delta(\varphi)+\varepsilon(\varphi)$$ (Note that making the identification concrete really paid off.)