# The Unapologetic Mathematician

## 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.

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Posted by John Armstrong | Differential Topology, Topology

## 4 Comments	»

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 | 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 | 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 | 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 | August 21, 2012 | Reply

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