Consider $(\mathbb R, +)$ and $S^1 := \mathbb R / 2 \pi \mathbb Z$. Both of these Lie groups have Lie algebra $(\mathbb R, [\cdot,\cdot] = 0)$. Does there exists a Lie group morphism $\Phi: S^1 \to \mathbb R$ such that $(\Phi_*)_e = \text{id}_{\mathbb R}$, where $e$ denotes the identity element of $S^1$?
My idea is the following. As $\Phi: S^1 \to \mathbb R$ is a Lie group morphism, it is continuous. Hence, as $S^1$ is connected and compact, its image is a bounded, closed interval of $\mathbb R$. However, the only subgroup of $(\mathbb R, +)$ that is of the form of a bounded, closed interval is the trivial group, i.e. $[0, 0] = \{0\}$. Therefore, $\Phi \equiv 0$, so $(\Phi_*)_e \equiv 0$, which proves that there doesn't exist such a $\Phi$. Is it really that simple?