Differentiability on $\mathbb{R}^n$ under an equivalence relation. Let $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ be differentiable. Given an equivalence relation $\sim$ on $\mathbb{R}^n$, is $f:\mathbb{R}^n/\sim \rightarrow \mathbb{R}^n/\sim$ also differentiable?
The question that made me consider this is as follows:
Given the 2-sphere $\mathbb{S}^2$ (as a manifold), consider the stereographic projection defined in the usual way, e.g. from the 'South pole' as
\begin{align}
P_S:\mathbb{S}^2\setminus (0,0,-1) &\longrightarrow \mathbb{R}^2
\\ (x, y, z) &\longmapsto (\frac{x}{1+z}, \frac{y}{1+z}).
\end{align}
With a similarly defined maps $P_N$ for the 'North pole', this gives us an atlas on $\mathbb{S}^2$ (they are smoothly compatible charts), and so we can talk about differentiability.
Now define the equivalence relation for $x, x'\in \mathbb{S}^2$
\begin{align}
x\sim x' \iff x = -x.
\end{align}
Now we can define
\begin{align}
P_S':\mathbb{S}^2\setminus (0,0,-1)/\sim &\longrightarrow \mathbb{R}^2/\sim'
\\ [(x, y, z)] &\longmapsto [(\frac{x}{1+z}, \frac{y}{1+z})],
\end{align}
where given $(\frac{x}{1+z}, \frac{y}{1+z}), a\in \mathbb{R}^2$, define
\begin{align}
(\frac{x}{1+z}, \frac{y}{1+z}) \sim' a \iff a = (-\frac{x}{1-z}, -\frac{y}{1-z}).
\end{align}
Together with the analogous $P_N'$ (although I don't know if this is necessary any longer, since $(0,0,-1)\sim(0,0,1)$), this provides charts for  $\mathbb{S}^2/\sim$.
Assuming there is a function $f\in C^{\infty}(\mathbb{S}^2)$, then if $f(x) = f(-x) \forall x\in \mathbb{S}^2$, then it seems to me that this function would be well defined on $\mathbb{S}^2/\sim$. i.e. define $f'\in C^{\infty}(\mathbb{S}^2/\sim):[(x, y, z)] \mapsto f(x, y, z)$.
Is this correct? And if $f$ is differentiable on $\mathbb{S}^2$, will it be differentiable on $\mathbb{S}^2/\sim$?
 A: It depends on the equivalence relation. Differentiability is defined in smooth manifolds, and any quotient of a smooth manifold may not be another manifold. From what you say it seems that you have studied manifolds inmersed in $\mathbb{R}^n$, but this seems like it could be better understood by studying manifolds in an abstract setting. A reference for that is $\textit{Introduction to Smooth Manifolds}$, by Lee.
The result you are interested in can be found in the 2$^{nd}$ edition of the book as theorem 21.13 (9.13 in the 1$^{st}$ edition). Your equivalence relation is given by a discrete group acting freely and properly in $\mathbb{S}^2$ (see example 21.14), so the quotient is a manifold (diffeomorphic to the projective plane, $\mathbb{P}\mathbb{R}^2$).
This means that we have a smooth quotient map $\pi:\mathbb{S}^2\to\mathbb{S}/{\sim}$. In the case of discrete groups (like in your example), the map $\pi$ is a local diffeomorphism.
Now to answer the question, if a map $\mathbb{S}^2\to\mathbb{S}^2$ is smooth, then the map $\pi\circ f:\mathbb{S}^2\to\mathbb{S}/{\sim}$ is smooth (as it is a composition of smooth maps). Now as you say, the condition $f(x)=f(-x)\;\forall x$ means that from $\pi\circ f$ we can obtain a well defined function $g:\mathbb{S}/{\sim}\to\mathbb{S}/{\sim}$. Moreover smoothness can be checked locally, and locally the function $g$ is given by $f\circ\pi^{-1}$, which is differentiable because $\pi$ is a local diffeomorphism.
PS: $\textit{Introduction to Smooth Manifolds}$ is the main book I´ve used to study smooth manifolds but I don´t know if it´s the best reference for quotients of manifolds by group actions at this level, anyone with a better reference is welcome to edit the answer.
