I want to calculate the cohomolgy group of the real projective space $RP^n$ based on this link (page 3). So, as is done here let $A:\mathbb{S}^n\to\mathbb{S}^n$ given by $A(x)=-x$. Since $A^2=I_{\mathbb{S}^n}$ and if $\Omega^{k}(\mathbb{S}^n)$ is the set of $k$-forms on $\mathbb{S}^n$ then $A^{*}:\Omega^{k}(\mathbb{S}^n)\to \Omega^{k}(\mathbb{S}^n)$ the pull-back has the same property $(A^{*})^2=I$. Therefore we can split $\Omega^{k}(\mathbb{S}^n)=\Omega^{k}(\mathbb{S}^n)_{+}\oplus\Omega^{k}(\mathbb{S}^n)_{-}$ where $\Omega^{k}(\mathbb{S}^n)_+$ is the subspace correspoding with the eigenvalue $+1$ of $A^*$, and similarly to the other subspace.
Now, if $\pi:\mathbb{S}^n\to RP^n$ is the canonical projection then we can show that if $\pi^*:\Omega^{k}(RP^n)\to\Omega^{k}(\mathbb{S}^n)$ is its pull-back, then $\pi^*(\Omega^{k}(RP^n))\subset\Omega^{k}(\mathbb{S}^n)_{+}$. Now, in the same link above mentioned, the calculation of the groups is finished by showing that $\pi^*:\Omega^{k}(RP^n)\to\Omega^{k}(\mathbb{S}^n)_{+}$ is an isomorphism.
The proof given was:
Let $\eta\in\Omega^{k}(\mathbb{S}^n)$ be a $k$-form. Recall that $\pi^*$ is defined by $$(\pi^*\omega)_p(v_1,\dots)=\omega_{\pi(p)}((\pi_*)_{p}v_1,\dots)$$ Hence $\pi^*\omega=\eta$ if and only if $$\omega_q(\omega_1,\dots)=\eta_p(v_1,\dots) $$ for every $p\in\mathbb{S}^n$ such that $\pi(p)=q$ and $v_i \in\mathbb{S}^n_p$ such that $(\pi_*)_p v_i=\omega_i$. But for every $q,\omega_1,\dots$ there are exactly two choices of $p,v_1,\dots$ ; if $p,(\pi_{*p})^{-1}\omega_1$ is one, then $A(p), A_{*}(\pi_{*p})^{-1}\omega_1$ is the other. Hence the last formula above determines $\omega$ uniquely if and only if $$\eta_{p}(v_1,\dots)=\eta_{A(p)}(A_{*}v_{1},\dots)=(A^{*}\eta)(v_1,\dots)$$ For all $p,v_{1},\dots$. i.e. if and only if $\eta\in\Omega^{k}(\mathbb{S}^{n})_{+}$.
However I could not understand in a clear way that this indeed assure the claim that $\pi^{*}$ is an isomorphism. I know that it is a basic logic but I'm having some difficults to see this. Can anyone explain me in a more clear way? Thanks!