Let $X$ be a smooth manifold of dimension $n > 1$. Let us denote by $\underline{\mathbb{S}}^{1}$ the sheaf of the smooth functions over circle, $C^{\infty}$ the sheaf of the smooth functions over $\mathbb{R}$ and by $\underline{\mathbb{Z}}$ the sheaf of the locally constant functions over $\mathbb{Z}$. So, we have an exact sequence of sheaves


$$0 \longrightarrow \underline{\mathbb{Z}} \stackrel{2\pi}\longrightarrow C^{\infty} \stackrel{exp}\longrightarrow \underline{\mathbb{S}}^{1} \longrightarrow 1,$$
where the map $exp: C^{\infty} \rightarrow \underline{\mathbb{S}}^{1}$ is $f \mapsto e^{2\pi i f}$ and the $2\pi$ map is simply multiplication by $2\pi$. Given any open $U \subset X$, it seems to me that the sequence of Abelian groups


$$0 \longrightarrow \underline{\mathbb{Z}}(U) \longrightarrow C^{\infty}(U) \longrightarrow \underline{\mathbb{S}}^{1}(U) \longrightarrow 1,$$
is an exact. But I saw in a pdf about sheaves that we need to ask that $U$ be simply connected. Is that correct? Is there any element in $\operatorname{ker}(exp)$ that is not in the image $\operatorname{Im}(2\pi)$?



Exact sequence of sheaves that generates an exact sequence of Abelian groups
Not sure the question belongs to MO but here you go : if $X = \mathbb C^\*$, consider $f = \frac{z}{|z|} \in C^{\infty}(X, S^1)$. If $f$ is in the image of $exp$, then we found a determination of the logarithm on $S^1$ which is impossible by classical complex analysis.


