Let $f\colon S^{11}\to S^6$ be a smooth map, it Hopf invariant is defined as $$ H(f)=\int_{S^{11}}\varphi\wedge\mathrm{d}\varphi $$ where $\varphi$ is a $5$-form such that $\mathrm{d}\varphi=f^*\alpha$, and $\alpha$ is a closed $6$-form such that $\int_{S^6}\alpha=1$
Now I'm going to show $H(f)$ must be even, here is my proof:
Let $\rho$ denote the antipodal map on $S^{11}$, that is $x\mapsto-x$, set $$ \beta=\frac12(\rho^*\varphi-\varphi) $$ then $\varphi+\beta$ is $\rho^*$-invariant, thus there exists a $5$-form $\omega$ on $\mathbb{RP}^{11}$ such that $\varphi+\beta=\pi^*\omega$, where $\pi\colon S^{11}\to\mathbb{RP}^{11}$ is canonical projection. Furthermore, since $\rho$ is homotopic to $\operatorname{id}_{S^{11}}$, then $\beta$ is exact, thus $$ \begin{aligned} \int_{S^{11}}\varphi\wedge\mathrm{d}\varphi&=\int_{S^{11}}(\varphi+\beta)\wedge\mathrm{d}(\varphi+\beta)\\ &=2\int_{\mathbb{RP}^{11}}\omega\wedge\mathrm{d}\omega \end{aligned} $$ Now it remains to show $$ \int_{\mathbb{RP}^{11}}\omega\wedge\mathrm{d}\omega\in\mathbb{Z} $$
I think it must be related to the condition $\int_{S^6}\alpha=1$, but I don't know how to use it. Is there anyone can give me some help? Thanks in advance!