The Unapologetic Mathematician

Mathematics for the interested outsider

Pullbacks on Cohomology

We’ve seen that if f:M\to N is a smooth map of manifolds that we can pull back differential forms, and that this pullback f^*:\Omega(N)\to\Omega(M) is a degree-zero homomorphism of graded algebras. But now that we’ve seen that \Omega(M) and \Omega(N) are differential graded algebras, it would be nice if the pullback respected this structure as well. And luckily enough, it does!

Specifically, the pullback f^* commutes with the exterior derivatives on \Omega(M) and \Omega(N), both of which are (somewhat unfortunately) written as d. If we temporarily write them as d_M and d_N, then we can write our assertion as f^*(d_N\omega)=d_M(f^*\omega) for all k-forms \omega on N.

First, we show that this is true for a function \phi\in\Omega^0(N). It we pick a test vector field X\in\mathfrak{X}(M), then we can check

\displaystyle\begin{aligned}\left[f^*(d\phi)\right](X)&=\left[d\phi\circ f\right](f_*(X))\\&=\left[f_*(X)\right]\phi\\&=X(\phi\circ f)\\&=\left[d(\phi\circ f)\right](X)\\&=\left[d(f^*\phi)\right](X)\end{aligned}

For other k-forms it will make life easier to write out \omega as a sum

\displaystyle\omega=\sum\limits_I\alpha_Idx^{i_1}\wedge\dots\wedge dx^{i_k}

Then we can write the left side of our assertion as

\displaystyle\begin{aligned}f^*\left(d\left(\sum\limits_I\alpha_Idx^{i_1}\wedge\dots\wedge dx^{i_k}\right)\right)&=f^*\left(\sum\limits_Id\alpha_I\wedge dx^{i_1}\wedge\dots\wedge dx^{i_k}\right)\\&=\sum\limits_If^*(d\alpha_I)\wedge f^*(dx^{i_1})\wedge\dots\wedge f^*(dx^{i_k})\\&=\sum\limits_Id(f^*\alpha_I)\wedge f^*(dx^{i_1})\wedge\dots\wedge f^*(dx^{i_k})\\&=\sum\limits_Id(\alpha_I\circ f)\wedge f^*(dx^{i_1})\wedge\dots\wedge f^*(dx^{i_k})\end{aligned}

and the right side as

\displaystyle\begin{aligned}d\left(f^*\left(\sum\limits_I\alpha_Idx^{i_1}\wedge\dots\wedge dx^{i_k}\right)\right)&=d\left(\sum\limits_I(\alpha_I\circ f)f^*(dx^{i_1})\wedge\dots\wedge f^*(dx^{i_k})\right)\\&=d\left(\sum\limits_I(\alpha_I\circ f)d(f^*x^{i_1})\wedge\dots\wedge d(f^*x^{i_k})\right)\\&=\sum\limits_Id(\alpha_I\circ f)\wedge d(f^*x^{i_1})\wedge\dots\wedge d(f^*x^{i_k})\\&=\sum\limits_Id(\alpha_I\circ f)\wedge f^*(dx^{i_1})\wedge\dots\wedge f^*(dx^{i_k})\end{aligned}

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 \Omega^k(N) and \Omega^k(M). And then immediately we get homomorphisms H^k(N)\to H^k(M), which we also write as f^*.

If you want, you can walk the diagrams yourself to verify that a cohomology class in H^k(N) is sent to a unique, well-defined cohomology class in H^k(M), but it’d probably be more worth it to go back to read over the general proof that chain maps give homomorphisms on homology.

July 21, 2011 Posted by | Differential Topology, Topology | 5 Comments