Smoothness of function I would like to describe the smoothness of the function over $S^2$ as a manifold, giving hours of daylight as a function of latitude. For simplicity, we just assume that the sun's direction is fixed.
To analyse this, I know I need to look at the smoothness of the function composed with a set of coordinate charts of $S^2$. I use two coordinate charts, the inverse of the north and south stereo-graphic projections. Which are given in spherical coordinates (here for the inverse projection from the North):
$F: (\cot(\frac{\psi}{2}), \cos \theta) = (\frac{2\cot(\frac{\psi}{2})cos\theta)}{\cot^2(\frac{\psi}{2})+1}, \frac{2\cot(\frac{\psi}{2})sin\theta)}{\cot^2(\frac{\psi}{2})}, \frac{1-\cot^2(\frac{\psi}{2})}{1+\cot^2(\frac{\psi}{2})})$
The hours of daylight function, I have derived from the "Sunrise equation": https://en.wikipedia.org/wiki/Sunrise_equation 
So the hours of daylight function is undefined at this latitude, and hence not smooth as a function over the manifold $S^2$.
Is my logic sound here? I'm trying to intuitively picture what may be happening at these points to make this function undefined, but I am struggling to visualize this.
 A: Let's begin at the beginning. Let $H : S^2 \to \mathbb{R}$ denote our would-be function giving the hours of daylight at each point on $S^2$, so that by the sunrise equation,
$$
 \cos\left(\frac{\Omega}{2}H(\mathbf{r})\right) := -\tan(\phi(\mathbf{r}))\tan(\delta),
$$
where $\phi : S^2 \to [0,\pi]$ is the latitude function, defined by
$$
 \phi(\mathbf{r}) := \arccos(z),
$$
where $\delta$ is the sun's declination, and where $\Omega$ is the angular velocity of the earth; in particular, given the maximal value of the sun's declination, we have $\lvert \delta \rvert \leq 0.13021823 \pi < \pi/4$. Note that if $\delta = 0$, which corresponds to the vernal or autumnal equinox, then we necessarily get
$$
 H(\mathbf{r}) = \frac{2}{\Omega} \cdot \left(k+\frac{1}{2}\right)\pi = \left(k+\frac{1}{2}\right)\frac{2\pi}{\Omega}
$$
for some $k \in \mathbb{N}$, where observation, of course, gives us $k=0$ and exactly half a day of daylight. So, let's restrict ourselves to the non-trivial case where $0 < \lvert\delta\rvert < \pi/4$.
Now, given that $\cos : \mathbb{R} \to [-1,1]$ and the fact that $0 < \lvert\delta\rvert < \pi/4$, the sunrise equation forces the condition
$$
 -\cot(\lvert\delta\rvert) \leq \tan(\phi(\mathbf{r})) \leq \cot(\lvert\delta\rvert),
$$
which, by a little trigonometric manipulation, is equivalent to
$$
 \frac{\pi}{2} - \lvert\delta\rvert \leq \phi(\mathbf{r}) \leq \frac{\pi}{2} + \lvert\delta\rvert.
$$
Thus, $H$ only actually makes sense as a function $\phi^{-1}\left(\left[\tfrac{\pi}{2}-\lvert\delta\rvert,\tfrac{\pi}{2}+\lvert\delta\rvert\right]\right) \to \mathbb{R}$, in which case
$$
 H(\mathbf{r}) = \frac{2}{\Omega}\arccos\left(-\tan(\phi(\mathbf{r}))\tan(\delta)\right);
$$
in particular, this should make you expect trouble at the latitudes $\phi^{-1}\left(\tfrac{\pi}{2}-\lvert\delta\rvert\right)$ and $\phi^{-1}\left(\tfrac{\pi}{2}+\lvert\delta\rvert\right)$, the boundary between the annular neighbourhood $\phi^{-1}\left(\left(\tfrac{\pi}{2}-\lvert\delta\rvert,\tfrac{\pi}{2}+\lvert\delta\rvert\right)\right)$ of the equator with distinct night and day and the regions of 24-hour daytime or nighttime. Indeed, since $\phi$ is smooth on
$$
 \phi^{-1}\left(\left(\tfrac{\pi}{4},\tfrac{3\pi}{4}\right)\right) \supset \phi^{-1}\left(\left[\tfrac{\pi}{2}-\lvert\delta\rvert,\tfrac{\pi}{2}+\lvert\delta\rvert\right]\right),
$$
it follows that $H$ fails to be differentiable precisely on the latitudes at the latitudes $\phi^{-1}\left(\tfrac{\pi}{2}-\lvert\delta\rvert\right)$ and $\phi^{-1}\left(\tfrac{\pi}{2}+\lvert\delta\rvert\right)$, precisely because $\arccos : [-1,1] \to [0,\pi]$ fails to be differentiable at $\pm 1$. Explicitly, on the domain $U := \phi^{-1}\left(\left[\tfrac{\pi}{2}-\lvert\delta\rvert,\tfrac{\pi}{2}+\lvert\delta\rvert\right]\right)$, we can safely apply the chain rule (and a bunch of trigonometric identities) to conclude that for any vector field $X$ defined on $U$ (e.g., $X = \tfrac{\partial}{\partial \phi}$ constructed via spherical coordinates), we have
$$
 D_X H(\mathbf{r}) = \operatorname{sign}\left(\cos\left(\phi(\mathbf{r})\right)\right)\frac{2}{\Omega}\frac{\sin(\delta)}{\sqrt{\cos(\phi+\delta)\cos(\phi-\delta)}}D_X\phi(\mathbf{r}),
$$
so that for suitable $X$ (e.g., $X = \tfrac{\partial}{\partial \phi}$), we have that $D_X H(\mathbf{r}) \to \pm \infty$ or $\mp\infty$ (depending on the signs of $\delta$ and $\cos\left(\phi(\mathbf{r})\right)=z$) as $\mathbf{r}$ approaches the latitude $\phi^{-1}\left(\tfrac{\pi}{2}\pm\lvert\delta\rvert\right)$. Again, this failure of differentiability occurs precisely at the boundary between the interior of the domain of $H$, which represents the region with distinct day and night, and the regions with 24-hour daytime or nighttime.
