I am trying to prove that:
The real projective space $\mathbb R\mathrm{P}^n$ is orientable if and only if $n$ is odd.
For do so, consider first the antipode map $\sigma:\mathbb R^{n+1}\to \mathbb R^{n+1}$, $p\mapsto -p$ and denote $\sigma^\star$ its pullback. The $(n-1)$-form in $\mathbb R^{n+1}$: $$\Omega=\sum_{\alpha=1}^{n+1}(-1)^{\alpha+1}\, dx^1\wedge\ldots\wedge dx^{\alpha-1}\wedge dx^{\alpha+1}\wedge \ldots \wedge dx^{n+1}\qquad\qquad\qquad (\star)$$ induces a volume form in $S^n\subset\mathbb R^{n+1}$ and satisfies $$\sigma^\star_p\Omega_p=(-1)^{n+1}\Omega_p$$ for each $p\in S^n$
Now, suppose that $\mathbb R\mathrm P^n$ is orientable, then there exists a volume form $\Lambda$ in $\mathbb R\mathrm P^n$. If $\pi$ is the quotient map $S^n\to\mathbb R\mathrm P^n$ then $\pi^\star \Lambda$ is a $n$-form in $S^n$ and there must exists a smooth map $f:S^n\to\mathbb R$ such that $$\pi^\star \Lambda=f\Omega$$ in $S^n$. Composing both sides of the last equation by $\sigma^\star$ and using $(\star)$ we obtain: $$\sigma^\star (\pi^\star \Lambda)=\sigma^\star (f\Omega)\Longleftrightarrow f\Omega=(-1)^{n+1}f\Omega$$ Since $f\ne 0$ in $S^2$ (why?) then $n$ must be odd.
I am not able to prove the converse. Any help?