# Map $f:M\to M$ such that $f^*\omega=-\omega$

This a test question from Jänich:

Let $M \neq\emptyset$ (a smooth manifold) and $1 \leq k \leq n=\dim M$. Can there exist a map $f:M\to M$ with the property that $f^*\omega=-\omega$ for all $\omega \in \Omega^kM$?

The answer key says NO, NEVER! But I don't know why, this is what I get if I suppose yes:

$\omega_{f(p)}(df_pv_1,\dots,df_pv_n)+\omega_p(v_1,\dots,v_n)=0$

Since this is must be true for all $\omega \in \Omega^kM$, I could reach a contradiction by using a particular $\omega$, but who?

• There is no such a thing as an n-form det on a manifold. – Mariano Suárez-Álvarez Mar 4 '18 at 3:08
• Consider the antipodal map $\rho:S^1 \to S^1$, and write out the action of $\rho$ on the trivial bundle $T^*S^1$. – anomaly Mar 4 '18 at 3:18