# Elementary proof of the third Lie theorem

I would like to understand a (not well known) proof due to G.M; Tuynman, of the third Lie theorem, which asserts that for any given finite dimensional Lie algebra $\mathcal{G}$ there exists a (simply connected) Lie group $G$ whose Lie algebra is $\mathcal{G}$.

The proof is similar to using the theorem of Ado, but requires some tools:

1) The notion of universal cover.

2) The fact that for a simply connected Lie group $G$ not only the first de Rham cohomology space $H^1(G)=\{0\}$ but also $H^2(G)=\{0\}$.

My problem is:

1) To acquire quickly the concepts of algebraic topology: simple connectedness, universal covering, fundamental group ...

2) Review the fact that $H^2(G)=\{0\}$ (unfortunately, I do not have the book of Godement). 3) Understanding the example discussed in the Tuynman's paper

http://ifile.it/hy0q139

I would like also to add, I get a copy , I find it excellent to deepen his knowledge of algebraic topology. I ran on page 202 of the Godbillon book. This is the theorem Künneth that identifies the cohomology of product manifolds with the tensor product of their cohomologies: $$K : H(M)\otimes H(N)\rightarrow H(M\times N)$$ If $M$ is compact, $K$ is an isomorphism. There are two things I can not understand, among others:
1) Godbillon consider $d=K^{-1}\circ\mu^\star$, where $\mu$ is the multiplication on the Lie group $G$, and $K$ is invertible, while the compactness is essential for $K$ to be an isomorphism?!
2) Godbillon theorem shows in his page 202, with $d$, if $G$ is connected then the smallest integer $q>0$ such that $H^q(G)\neq 0$ is odd. First, $G$ must be compact? Why $H^1(G)=0$? This is not a consequence of the above theorem where compactness is essential!