# 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.

July 21, 2011