Let \(F\colon M\to N\) be smooth. Since a differential form \(\omega\) is a tensor field , the pullback \(F^*\omega\) is defined by
\begin{equation*} (F^*\omega)_p(v_1, \ldots, v_k)=\omega_{F(p)}(dF_p(v_1),\ldots,dF_p(v_k)). \end{equation*}This is again a differential form.
Remarks
- \(F^*\colon \Omega^k(N)\to \Omega^k(M)\) is linear
- the pullback respects the wedge product , to be more precise, we have \begin{equation*} F^*(\omega\wedge \eta)=F^*(\omega)\wedge F^*(\eta). \end{equation*}