What is the differential of the quotient map? We can view the projective space $P(\mathbb R^n)$ as the quotient of $S^n/\sim$  where $x \sim y$ if and only if $x = -y$.
The quotient map $q: S^n \to P(\mathbb R^n)$ is the map $x \mapsto [x]$ where $[x] = \{x,-x\}$. 
I want to calculate the Jacobian matrixdifferential of $q$ but I don't see what it is. To me it seems that it is both the identity $I$ and $-I$ at the same time. 

What is the differential of $q$ and how to caluclate it?

Here is what I have so far:
Since I didn't know how to do the general case I tried to do the explicit caluclation for $n=2$. 
Let $\varphi : \mathbb R^2 \to S^2, (x,y) \mapsto ({2x \over x^2 + y^2 + 1}, {2y \over x^2 + y^2 +1}, {x^2 +y^2 -1\over x^2 +y^2 +1})$ and $\psi^{-1}:P(\mathbb R^2) \to \mathbb R^2, [(x,y,z)] \mapsto ({y\over x}, {z\over x})$ and $F: S^2 \to P(\mathbb R^2), x \mapsto [x]$.
Then 
$$ \psi^{-1}\circ F \circ \varphi (x,y) = ({y\over x}, {x^2 +y^2 -1 \over 2x}) $$
and the Jacobian I calculated as
$$ J_{\psi^{-1}\circ F \circ \varphi} =\left ( \begin{matrix} -{y \over x^2} & {1\over x} \\ {2x^2 - 2y^2 + 2 \over 4 x^2 } & {y\over x} \end{matrix}\right )$$

Is this correct? And if so, how can I generalise to arbitrary $n$?

 A: Sometimes, when doing calculations, it is convenient to view tangent vectors as velocities on smooth trajectories. Formally, if $x \in S^n$ and $v \in T_x S^n$, then there exist a smooth curve $\gamma : (-\varepsilon, \varepsilon) \to S^n$ with $\gamma (0) = x$ and $\dot \gamma (0) = v$.
The correct consequence of $q(x) = q(-x)$ is that
$$(\Bbb d _x q) (v) = (\Bbb d _x q) (\dot \gamma (0)) = \frac {\Bbb d} {\Bbb d t} (q (\gamma (t)) (0) = \frac {\Bbb d} {\Bbb d t} (q (\color{red} - \gamma (t)) (0) = (\Bbb d _{\color{red} - x} q) (\color{red} - \dot \gamma (0)) = (\Bbb d _{\color{red} - x} q) (\color{red} - v)$$
and this part should clarify the confusion about $\Bbb d q$ "being both $\text{Id}$ and $-\text{Id}$".
As to performing explicit computations, this is doubly impossible to do because:


*

*unlike on $\Bbb R^n$, there are no global coordinates neither on $S^n$, nor on $P^n(\Bbb R)$, so any computation would be local and the formulae would depend heavily on the concrete formulae of the charts used (as you have discovered yourself, above)

*at a more fundamental level, the points of $P^n(\Bbb R)$ are not defined via any explicit formula, but merely by the very formal and abstract formula $x \sim -x$, therefore you have no explicit formula for $[x]$ to work with; the equivalence class of $x$ is the equivalence class of $x$, that's it, unfortunately; you may squeeze some useful properties out of this (as I've done above), but don't expect explicit formulae.
Are you surprised? You shouldn't, because you've seen the same problem arise in a completely different and elementary context - but without realizing it. Can you explicitly compute $\sqrt 2$? You know how to define it (the positive root of $x^2 = 2$), but can you explicitly compute it? No, you can't (you can compute arbitrarily good approximations of it, but not it itself). Nevertheless, its definition enables you to do some manipulations with it, and conclude plenty of things about it, such that $(\sqrt 2)^4 = 4$, or that it is irrational - without really knowing its explicit form.
Adopting a very Galois-minded approach ("Sauter à pieds joints sur ces calculs"), I'd say that obtaining mathematical understanding through non-computational means (because most of the time they are impossible) is at the core of mathematics. You'll encounter this inner tension between the desire to compute (because computation is easy and algorithmic, and it doesn't require intellectual ingenuity and creative effort - and we all love to be lazy, don't we?) and the impossibility to do it in every branch of higher mathematics - save, maybe, for numerical analysis (which aims to compute). If you plan to pursue a research career in mathematics, the sooner you emotionally and mentally accept this, the better. Mathematical thinking is all about intelligently avoiding ipossible computations, about making a (sometimes long and strenuous) detour where a straight path is desirable but impossible.
As to your concrete question, it is essentially computational, therefore it doesn't fit well in the conceptual and operational framework of smooth manifolds. This is the kind of question that you should not ask in differential geometry. If it is part of a larger problem, then the answer of that problem is most surely non-computational, and you've chosen the wrong approach to solving it.

To answer the question that you raise in the first comment below this post: essentially, $q$ is a covering map (covering $P^n (\Bbb R)$ with two sheets), so it is a local diffeomorphism, and as such we may locally push-forward the local restrictions of $\omega$ and then show that their images are compatible with each other on $P^n (\Bbb R)$ and so give birth to a form $\omega'$ on $P^n (\Bbb R)$. The gory details follow.
Let $x \in S^n$ and choose an open $x \in U \subset S^n$ such that for every $y \in U$, $-y \notin U$. Let $q_U = q \big| _U$. Notice that $q_U : U \to q(U)$ is invertible, because if $p \in P^n (\Bbb R)$ we may define $q_U ^{-1} (p)$ to be that unique $y \in U$ such that $q (y) = p$ ($y$ is unique due to how we chose $U$). Let $s_U = q_U ^{-1}$. With this done, and since $q_U$ is smooth, it can be shown that $q_U : U \to q(U)$ is a diffeomorphism, so $q_U ^{-1} : q(U) \to U$ will be a diffeomorphism too.
Take now $\omega \in \Omega^1 (S^n)$ and define $\omega' _U = s_U ^* (\omega \big| _U) \in \Omega^1 (q(U))$. This is not a form on $P^n (\Bbb R)$, but only on the open subset $q(U)$. Fortunately, each $x \in S^n$ admits a neighbourhood $U_x$ as above, so these neighbourhoods will cover $S^n$, so their images $\{q(U_x)\} _{x \in S^n}$ will form an open cover of $P^n (\Bbb R)$. If we show that for any such two overlapping $q(U_x), q(U_y)$, the corresponding local forms $\omega' _{U_x}$ and $\omega' _{U_y}$ agree on the overlapping, then all these local forms $\{\omega' _{U_x}\} _{x \in S^n}$ will glue together into a single form $\omega' \in \Omega^1 (P^n (\Bbb R))$ that will be the desired form (and $\omega' _U = \omega' \big| _{q(U)}$).
For the simplicity of writing, let $V = q(U_x) \cap q(V_x)$ and keep in mind that
$$\omega_{U_x} \big| _{U_x \cap U_y} = \omega \big| _{U_x \cap U_y} = \omega_{U_y} \big| _{U_x \cap U_y}$$
and that
$$s_{U_x} \big| _V = q_{U_x} ^{-1} \big| _V = q_{U_y} ^{-1} \big| _V = s_{U_y} \big| _V$$
so that we may finally write
$$\omega' _{U_x} \big| _V = (s_{U_x} ^* \omega _{U_x}) \big| _V = (\omega _{U_x} \circ \textrm d s_{U_x}) \big| _V = \omega _{U_x} \big| _{s_{U_x} (V)} \circ \textrm d (s_{U_x} \big| _V) = \\
\omega _{U_y} \big| _{s_{U_y} (V)} \circ \textrm d (s_{U_y} \big| _V) = (\omega _{U_y} \circ \textrm d s_{U_y}) \big| _V = (s_{U_y} ^* \omega _{U_y}) \big| _V = \omega' _{U_y} \big| _V$$
so indeed the each such two local forms agree on the corresponding overlaps, thus being local restrictions of a unique form $\omega'$ which is the "push-down" of $\omega$.
A: I am afraid you are trying to get information in a situation where not a lot of information is available. The only "general" statement that you can make about the derivative is that it is always a linear isomorphism, everything else depends on the choice of coordinates. For the choice of local charts (or coordinates) you made, the computation you did basically looks OK, apart from the fact that you did not properly restrict the domains of definition. The chart you denoted by $\psi^{-1}$ is not defined on all of $P(\mathbb R^2)$ but only on the complement of a projective line. Denoting this open subset by $U$, you get $U=\{[(x,y,z)]:x\neq 0\}$. To compute the derivative, you should now restrict to $F^{-1}(U)\subset S^2$ and then to $\phi^{-1}(F^{-1}(U))=\{(x,y):x\neq 0\}$. Then the Jacobian makes sense as calculated (and the computation looks OK to me), and in particular, you see that it is always invertible. However, the explict expression completely depends on the charts you choose. 
For example, if you take $B$ to be the open unit ball in $\mathbb R^2$ and the local chart $\phi(x,y):=(x,y,\sqrt{1-x^2-y^2})$, then you can use $F\circ\phi$ as a chart on $P(\mathbb R^2)$, and the coordinate expression of $F$ in this chart is the identity, so also the Jacobian is the identity.   Likewise, you can use $\tilde\phi(x,y):=(x,y,-\sqrt{1-x^2-y^2})$ as a chart on $S^2$ and continue using $F\circ\phi$ as a chart on $P(\mathbb R^2)$. Then both $F$ and its derivative will become $-id$, and so on. 
A: Jacobian matrix are defined for differentiable maps $f:R^n\rightarrow R^m$, when for differentiable morphisms $f:M\rightarrow N$ you may use a similar notion which depends on charts. 
https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant
