Method to calculate the de Rham cohomology of $\mathbb{R}\mathrm{P}^n$ I'm trying to follow through a method to calculate the de Rham cohomology groups of $\mathbb{R}\mathrm{P}^n$ from the de Rham cohomology groups of $S^n$.
I'm trying to show that differential k-forms on $\mathbb{R}\mathrm{P}^n$ are in 1-1 correspondence with differential k-forms, $\alpha$ on $S^n$ that satisfy $\sigma^* \alpha = \alpha$, where $\sigma$ is the antipodal map. I've started trying to show this for the case where $\alpha$ is a 1-form as an example but I am having difficulty even there.
If we let $\pi: S^n \to \mathbb{R}\mathrm{P}^n$ be the map $x \to [x]$ (i.e. sends points on sphere to line through origin) then as this map is surjective, the derivative maps $T_x\pi$ and $T_{\sigma(x)}\pi$ are also surjective and are therefore isomorphisms. (I'm not sure that map is surjective implies derivative surjective so would be interested to see an argument here as well!). Now given a 1-form $\alpha$ with $\sigma^* \alpha = \alpha$ we can define a form on $\mathbb{R}\mathrm{P}^n$ by:
$$\beta([x]) = ((T_x\pi)^{-1})^* \alpha(x)$$
Now I'd like to show this is well-defined by showing that $\beta([x]) = \beta([-x])$. We see that $\beta([-x]) = ((T_{\sigma(x)}\pi)^{-1})^* \alpha(-x)$ but now I'm struggling to equate this to $\beta([x])$. Is it true that $\alpha(-x) = \sigma^* \alpha(x)$ and given that how do we make a relation between $((T_{\sigma(x)}\pi)^{-1})^* \sigma^*$ and $((T_x\pi)^{-1})^*$? It's also not clear to me how we can show this correspondence is 1-1?
Once I have sorted this issue I would like to then find the cohomology groups. We can define a projection the k-forms on $S^n$ by $\beta \to \frac{\beta + \sigma^*\beta}{2}$ and I believe this descends to a surjective map on the cohomology groups $H^k(S^n) \to H^k(\mathbb{R}\mathrm{P}^n)$ but then it seems the most I can say is that $H^k(\mathbb{R}\mathrm{P}^n)$ is a sub vector space of $H^k(S^n)$. 
Thank you for any help.
 A: First of all, since $\pi = \pi\circ\sigma$, if $\beta$ is a $k$-form on $\Bbb RP^n$, then $\alpha = \pi^*\beta = (\pi\circ\sigma)^*\beta = \sigma^*(\pi^*\beta) = \sigma^*\alpha$. Conversely, we know that $\pi$ is a local diffeomorphism. So, given a point $[x]\in\Bbb RP^n$, choose a neighborhood $U$ of $x\in S^n$ so that $\pi|_U\colon U\to \pi(U)$ is a diffeomorphism. Given a $k$-form $\alpha$ on $S^n$ satisfying $\alpha=\sigma^*\alpha$, on $\pi(U)$ we define the $k$-form $\beta  = \big((\pi|_U)^{-1}\big)^*\alpha$. (This is making local what you did at a point in your write-up.) Since $\pi|_{\sigma(U)} =\pi|_U\circ\sigma$, we have $(\pi|_{\sigma(U)})^{-1} = \sigma^{-1}\circ(\pi|_U)^{-1}=\sigma\circ(\pi|U)^{-1}$ and so it follows that $$ \big((\pi|_{\sigma(U)})^{-1}\big)^*\alpha = 
\big(\sigma\circ(\pi|U)^{-1}\big)^*\alpha = \big((\pi|U)^{-1}\big)^*\sigma^*\alpha = \big((\pi|U)^{-1}\big)^*\alpha;$$
thus, $\beta$ is in fact well-defined.
Thus, the vector space of $k$-forms on $\Bbb RP^n$ is the vector space of $\sigma$-invariant $k$-forms on $S^n$. Moreover, this equality respects closedness and exactness. Thus, $H^k(\Bbb RP^n) \cong H^k_\sigma (S^n)$, where by the latter we mean the vector space of closed $\sigma$-invariant $k$-forms mod exact $\sigma$-invariant $k$-forms. All that remains for you to do is to decide what $H^k_\sigma (S^n)$ is. The only interesting case is $k=n$, and I'll let you work on that for a while.
