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


  1. […] spaces are all contravariant functors on the category of smooth manifolds. We’ve even seen how it acts on smooth maps. All we really need to do is check that it plays nice with […]

    Pingback by De Rham Cohomology is Functorial « The Unapologetic Mathematician | July 23, 2011 | Reply

  2. […] where each term omits exactly one of the basic -forms. Since everything in sight — the differential operator and both integrals — is -linear, we can just use one of these terms. And so we can calculate the pullbacks: […]

    Pingback by Stokes’ Theorem (proof part 1) « The Unapologetic Mathematician | August 18, 2011 | Reply

  3. […] the exterior derivative — gives us a chain complex. Since pullbacks of differential forms commute with the exterior derivative, they define a chain map between two chain […]

    Pingback by The Poincaré Lemma (setup) « The Unapologetic Mathematician | December 2, 2011 | Reply

  4. […] The exterior derivative is a derivative, The exterior derivative is nilpotent, De Rham Cohomology, Pullbacks on Cohomology, De Rham cohomology is functorial, The Interior […]

    Pingback by Tenth Linkfest | August 21, 2012 | Reply

  5. “and luckily enough, it does” … hmmm, interesting way of looking at this–as “lucky”.

    Actually the fact is that if it did NOT, mathematicians would not be interested in THIS theory–so not “lucky” but necessary

    Comment by John rood | January 5, 2018 | Reply

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: