We have this:
Theorem 1. If $H$ and $G$ are Lie groups with $H$ simply connected, then for any Lie algebra homomorphism $\psi: h \rightarrow g$ there exists a unique Lie group homomorphism $\phi : H \rightarrow G $ with $d\phi = \psi$.
My question is how to find example of two Lie groups $H, G$ and a Lie algebra homomorphism $\psi: h \rightarrow g$ such that there exists no Lie group homomorphism $\phi : H \rightarrow G $ with $d\phi = \psi$?
I know that H can't be simply connected, but I have no idea for an example and how to prove in that case that does not exist such $\phi$?
For instance, take $H=S^1$, $G=\Bbb R$, and let $\psi\colon\mathfrak h(=\Bbb R)\longrightarrow\mathfrak g(=\Bbb R)$ be the identity map. Since the only continuous group homomorphism from $S^1$ into $\Bbb R$ is the trivial homomorphism, there is no homomorphism $\phi\colon S^1\longrightarrow\Bbb R$ such that $\operatorname d\phi=\psi$.