Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 115 / 2 / b

The de Rham cohomology of $X$ isBecause pullback commutes with $d$, it sends closed forms to closed forms and exact forms to exact forms. Thus a smooth map induces

Let $H:X\times I\to Y$ be a smooth homotopy and write $H_t(x)=H(x,t)$. If $V={\partial }_t$, Cartan's magic formula givesIntegrating defines a degree-minus-one operator $K$ satisfyingFor closed $\omega$, the difference is exact, so smoothly homotopic maps induce the same map on de Rham cohomology.

If $F:X\to Y$ is a homotopy equivalence with inverse up to homotopy $G$, functoriality and homotopy invariance giveHence $F^{\ast }$ is an isomorphism.