If $F:G\to H$ is a homomorphism of Lie groups then there exists a linear map ${F}_{*}:\U0001d524\to \U0001d525$ which is a Lie algebra homomorphism (i.e. ${F}_{*}[X,Y]=[{F}_{*}X,{F}_{*}Y]$ ) such that $F(\mathrm{exp}X)=\mathrm{exp}({F}_{*}X)$ for all $X\in \U0001d524$ .
Lie's theorem on homomorphisms
Exponentiating homomorphisms
Last time we proved the following theorem.
What about the other direction? If we're given a Lie algebra homomorphism $f:\U0001d524\to \U0001d525$ , can we find a Lie group homomorphism $F:G\to H$ such that ${F}_{*}=f$ ?
Take $G=H=U(1)$ . Then $\U0001d524=\U0001d525=i\mathbf{R}$ so, for any smooth homomorphism $F:U(1)\to U(1)$ we get ${F}_{*}:i\mathbf{R}\to i\mathbf{R}$ . Since ${F}_{*}$ is linear and $i\mathbf{R}$ is 1dimensional, ${F}_{*}(i\theta )=i\lambda \theta $ for some $\lambda \in \mathbf{R}$ .
The Lie bracket on $i\mathbf{R}$ is trivial: $[i{\theta}_{1},i{\theta}_{2}]={i}^{2}{\theta}_{1}{\theta}_{2}{i}^{2}{\theta}_{2}{\theta}_{1}=0$ , so ${F}_{*}$ automatically preserves the bracket. Therefore any $\lambda $ defines for us a Lie algebra homomorphism $i\mathbf{R}\to i\mathbf{R}$ , $i\theta \mapsto i\lambda \theta $ .
So any homomorphism $F:U(1)\to U(1)$ has the form $F({e}^{i\theta})={e}^{i\lambda \theta}$ for some $\lambda $ . Which $\lambda $ can occur? Not all of them. For $F$ to be welldefined, we need $F({e}^{i2\pi})=F(1)$ (because ${e}^{i2\pi}=1$ ) so we need ${e}^{i2\pi \lambda}=1$ and hence we need $\lambda \in \mathbf{Z}$ . Conversely, if $\lambda \in \mathbf{Z}$ then $F({e}^{i\theta})={e}^{i\lambda \theta}$ is welldefined: it's just raising ${e}^{i\theta}$ to the (integer) power $\lambda $ .
So not every Lie algebra homomorphism can be exponentiated to get a Lie group homomorphism: $i\theta \mapsto i\lambda \theta $ can be exponentiated if and only if $\lambda \in \mathbf{Z}$ .
This example is sufficiently important to summarise as a lemma:
The maps ${F}_{n}({e}^{i\theta})={e}^{in\theta}$ , $n\in \mathbf{Z}$ , are homomorphisms $U(1)\to U(1)$ and every smooth homomorphism $U(1)\to U(1)$ is one of these.
The integers come into this example because the group $U(1)$ has some interesting topology. $U(1)$ is a circle, so loops in $U(1)$ have a winding number that counts how many times they go around the circle. The $\mathbf{Z}$ in the lemma is related to this integer winding number.
Simplyconnectedness
There is a topological condition you can put on $G$ to ensure that any Lie algebra homomorphism exponentiates.
Let $G$ be a simplyconnected matrix group and $H$ any matrix group. Then for every Lie algebra homomorphism $f:\U0001d524\to \U0001d525$ there exists a smooth group homomorphism $F:G\to H$ such that ${F}_{*}=f$ .
We will not prove this theorem: it's a gorgeous proof and would make an excellent project!
A space $X$
is called
A loop in $X$ is a continuous map $\gamma :[0,1]\to X$ such that $\gamma (0)=\gamma (1)$ . You should think of $\gamma (t)$ as a parametrised path with parameter $t\in [0,1]$ , which starts and ends at the same point. If $X$ is a matrix group then continuity here means that the matrix entries of $\gamma (t)$ depend continuously on $t$ .
A nullhomotopy of a loop $\gamma $ is a continuous map $H:[0,1]\times [0,1]\to X$ (which we can think of as a family of loops ${\gamma}_{s}(t)=H(s,t)$ ) such that:

these loops all start and end at the same point $x$ (so $H(s,0)=H(s,1)=x$ for all $s\in [0,1]$ .

${\gamma}_{0}(t)=H(0,t)$ is our loop $\gamma (t)$

${\gamma}_{1}(t)=H(1,t)$ is the constant loop $H(1,t)=x$ .
In other words, ${\gamma}_{s}(t)$ starts being $\gamma $ and gradually shrinks down to a constant loop.
The circle is certainly not simplyconnected, but there are plenty of groups which are simplyconnected. There is an optional video which talks more about the topology of the groups we've been meeting which will give a vague explanation for why the groups $SU(n)$ are simplyconnected. I also have some videos about topology, simplyconnectedness and the fundamental group if you want to learn more about this (you could do a project about the topology of Lie groups).