(The image corresponds to the book "Lie Groups, Lie Algebras, and their Representations" by Varadarajan)
Hello. I have a question regarding Lie groups and Lie algebras. If I have a Lie algebra $\mathfrak{g}$ the theorem mentions that there is a simply connected Lie group $G$ such that its Lie algebra is isomorphic to $\mathfrak{g}$.
Question 1. Is such a group $G$ related to the exponential application? i.e. $G=\left\{\exp(X):X\in\mathcal{G}\right\}$?
(All Lie algebras are finite-dimensional here.) Generally:
These remarks apply in particular to the unique simply connected Lie group with Lie algebra $\mathfrak{g}$. However, generally it's not true that the exponential map is surjective.
Edit: Maybe I've misunderstood your question. Maybe you were asking: can one construct the unique simply connected Lie group $G$ with Lie algebra $\mathfrak{g}$ using the exponential map somehow?
The answer is not obviously yes, since in the absence of a specific Lie group it's not clear what "the exponential map" ought to mean. We can do something like this by first using Ado's theorem to find a faithful representation $\mathfrak{g} \to \mathfrak{gl}(V)$. This means we can apply the ordinary matrix exponential to $\mathfrak{g} \subset \mathfrak{gl}(V)$ which will generate a subgroup of $GL(V)$. This group is not necessarily simply connected, but it does have Lie algebra $\mathfrak{g}$ (this is not obvious but is standard) and we can take its universal cover.
As the comments above already point out, if $G$ is any Lie group with Lie algebra $\mathfrak g$, then there is an exponential map $\text{exp}_G\colon \mathfrak g \to G$ whose image generates $G^0$, the connected component of $G$ containing the identity element of $G$. This map is in general neither injective nor surjective however.
The failure of injectivity is easy to see already for $S^1$ the circle, and it is not too hard to see that the exponential map $\text{exp}\colon \mathfrak{sl}_2(\mathbb R) \to \mathrm{SL}_2(\mathbb R)$ from the Lie algebra $\mathfrak{sl}_2(\mathbb R)= \{X\in \text{Mat}_2(\mathbb R): \text{tr}(X)=0 \}$) to the special linear group $\mathrm{SL}_2(\mathbb R) = \{A\in \text{Mat}_2(\mathbb R): \det(A)=1\}$ which is given by $\text{exp}(X) = \sum_{k=0}^\infty \frac{X^k}{k!}$ is not surjective.
None of the above involves the fundamental group of $G$ however. Where the issue of simple-connectedness does naturally arise in the context of Lie's theorems, is when one asks about homomorphisms: if $\phi\colon \mathfrak g\to \mathfrak h$ is a Lie algebra homomorphism, and $\mathfrak g = \text{Lie}(G)$, $\mathfrak h = \text{Lie}(H)$, then it is natural to ask if one can "integrate" $\phi$ to a homomorphism of Lie groups $\Phi\colon G \to H$, such that $$ \require{AMScd} \begin{CD} \mathfrak{g} @>{\phi}>> \mathfrak h\\ @V{\exp_G}VV @VV{\exp_H}V\\ G @>{\Phi ?}>> H \end{CD} $$ Now since $\exp\colon \mathfrak g\to G$ is locally a diffeomorphism from $0_{\mathfrak g}$ to $e_G$ and the exponential of a derivation is a homomorphism, one expects $\phi$ to "locally lift" to a homomorphism from $G$ to $H$ -- indeed it must be given by $\exp_H(\phi(\log_G(g))$ where $\log$ is a local inverse to $\exp_G$ at $e_G$ -- but in general the ability to lift locally will not ensure the existence of a globally defined homomorphism from $G$ to $H$. Perhaps the simplest example of this global failure is given by taking $G=S^1$ and $H= \mathbb R$ -- then $\mathfrak g = \mathrm{Lie}(G)=\mathrm{Lie}(H) = \mathfrak h$, with both being equal to $\mathfrak{gl}_1(\mathbb R)$, that is, the real numbers as a Lie algebra with the zero Lie bracket. It follows that we have an obvious homomorphism from $\phi\colon \mathfrak g \to \mathfrak h$ given by the identity map, but there is no corresponding homomorphism $\Phi\colon S^1\to \mathbb R$ for it to integrate to (the image of $S^1$ under a continuous map will be compact, but $(\mathbb R,+)$ has no compact subgroups other than the trivial subgroup, hence the only continuous homomorphism from $S^1$ to $\mathbb R$ is the constant map to $0_{\mathbb R}$. On the other hand, if we let $G$ and $H$ trade places, then the exponential map itself gives the lift of $\phi^{-1}$ to a homomorphism of groups from $\mathbb R$ to $S^1$.
The crucial difference between the two groups turns out to be the fact that $\mathbb R$ is simply connected: whenever $G$ is simply connected, using the fact that a path can then be lifted from $\mathfrak g$ to $G$ uniquely up to homotopy, one can show that any $\phi\colon \mathfrak g\to \mathfrak h$ lifts to a group homomorphism $\Phi\colon G \to H$. (Note that, as we saw in the example of $\mathbb R^1$ and $S^1$, the group $H$ is not required to be simply-connected here, only $G$.)
