Let $\varphi,\psi:G\to H$ be Lie group homomorphisms, with G connected, such that the induced Lie algebra homomorphisms $d\varphi,d\psi:g\to h$ are identical.
I want to show that then $\varphi=\psi$. A proof can be found in Frank Warner's book, but it uses differential forms and I would like to prove it somewhat more directly. Any ideas?
Also, what if $G$ is not connected?
For any $X\in\mathfrak{g}$, the map $e_X:\mathbb{R}\to G$ given by $e_X(t)=\exp(tX)$ is a homomorphism, and is the unique homomorphism $\mathbb{R}\to G$ such that $e_X'(0)=X$. Note that $\varphi\circ e_X$ is a homomorphism $\mathbb{R}\to H$ whose derivative at $0$ is $d\varphi(X)$, so it must be $e_{d\varphi(X)}$. It follows that $\varphi(\exp(X))=\exp(d\varphi(X))$ for each $X\in\mathfrak{g}$, and similarly for $\psi$. Since $d\varphi=d\psi$, this means that $\varphi=\psi$ on the entire image of $\exp:\mathfrak{g}\to G$. But the image of $\exp$ contains a neighborhood of the identity, which generates the entire group since $G$ is connected. Since the set of $g\in G$ such that $\varphi(g)=\psi(g)$ is a subgroup, this implies $\varphi=\psi$ on all of $G$.
The connectedness hypothesis is absolutely necessary here. For instance, if $G$ is $0$-dimensional, then $d\varphi=d\psi$ tells you nothing at all.