Do Spherical harmonics have continuous extensions to the entire sphere?

Question

This article contains the following formula for the spherical harmonics: $$Y_l^m(\theta,\phi) = \sqrt{\frac{(2l+1)}{4\pi}\frac{(l-m)!}{(l+m)!}} P_l^m(\cos \theta) e^{im\phi}$$ Now let $S$ be the unit sphere in three dimensions. Clearly, the function\begin{align}F\colon\mathbb R\times\mathbb R&\to S\\(\theta,\phi)&\mapsto\begin{pmatrix}\sin\theta\cos\phi\\\sin\theta\sin\phi\\\cos\theta\end{pmatrix}\end{align} to the sphere is surjective, but it is necessary that $F(\theta,\phi)=F(x,y)$ implies that $Y(\theta,\phi)=Y(x,y)$ so that we can define $Y$ on $S$.

For example, consider the north pole $N=(0,0,1)$. Clearly, $F(0,\phi)=N$ for all $\phi\in\mathbb R$, but in general the function $$\mathbb R\ni\phi\mapsto Y(0,\phi)$$is not constant, is it?

Answer 1

For $m=0$, the term $e^{im\phi}$ is constant. For $m\neq 0$, the Legendre polynomial vanishes for $x=\pm1$

Answer 2

If we look at the formulae for spherical harmonics given in Wikipedia, we see that whenever the $m$ parameter is nonzero, the $\theta$ dependence contains factors of $\sin\theta$ (so it's not a polynomial in just $\cos\theta$), which multiply the nonconstant $\exp(im\phi)$ factor by zero at the poles. Thus no discontinuity exists.