Pullbacks on Cohomology
We’ve seen that if is a smooth map of manifolds that we can pull back differential forms, and that this pullback is a degree-zero homomorphism of graded algebras. But now that we’ve seen that and are differential graded algebras, it would be nice if the pullback respected this structure as well. And luckily enough, it does!
Specifically, the pullback commutes with the exterior derivatives on and , both of which are (somewhat unfortunately) written as . If we temporarily write them as and , then we can write our assertion as for all -forms on .
First, we show that this is true for a function . It we pick a test vector field , then we can check
For other -forms it will make life easier to write out as a sum
Then we can write the left side of our assertion as
and the right side as
So these really are the same.
The useful thing about this fact that pullbacks commute with the exterior derivative is that it makes pullbacks into a chain map between the chains of the and . And then immediately we get homomorphisms , which we also write as .
If you want, you can walk the diagrams yourself to verify that a cohomology class in is sent to a unique, well-defined cohomology class in , but it’d probably be more worth it to go back to read over the general proof that chain maps give homomorphisms on homology.