Finite point sources on an oriented circle

Put $U_N=X\setminus \{N\}$ and $U_S=X\setminus \{S\}$. These are open in the subspace topology and cover $X$. For the displayed coordinates $u={\sigma }_N(x,y)$ and $v={\sigma }_S(x,y)$, direct substitution gives ${\sigma }_N^{-1}(u)=(\frac{2u}{1+u^2},\frac{u^2-1}{1+u^2}),{\sigma }_S^{-1}(v)=(\frac{2v}{1+v^2},\frac{1-v^2}{1+v^2})\mathrm{.}$ Their denominators are positive; the first image misses $N$, the second misses $S$, and both pairs of compositions are identities. The maps and inverses are continuous, so they are charts onto $\mathbb{R}$. Since $X\subset {\mathbb{R}}^2$ is Hausdorff and second-countable, these charts exhibit a topological $1$-manifold. On $U_N\cap U_S$, each coordinate image is $\mathbb{R}\setminus \{0\}$, and both transitions are $w\mapsto 1\mathrm{/}w$, smooth there with smooth inverse. Thus [F3] gives exactly the smooth structure specified in the Example. The field $V$ has chart components $(1+u^2)\mathrm{/}2$ and $-(1+v^2)\mathrm{/}2$, respectively, so is smooth and nowhere zero and gives the orientation of [F4]. In a chart centered at a point, its singleton is the coordinate zero set, hence a compact embedded zero-dimensional submanifold with the stated positive orientation.

By [F1] and linearity, $\rho (f)=a{\delta }_{y_1}(f)+b{\delta }_{y_2}(f)=af(y_1)+bf(y_2)$; each evaluation is well-defined by [F2].

Taking $f=1$ gives $\rho (1)=a+b$ because $1(y_1)=1(y_2)=1$. If $y_1=y_2$, the same formula combines the two coefficients, so the displayed current equality follows.