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
&=\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}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cbegin%7Baligned%7D%5Cleft%5Bf%5E%2A%28d%5Cphi%29%5Cright%5D%28X%29%26%3D%5Cleft%5Bd%5Cphi%5Ccirc+f%5Cright%5D%28f_%2A%28X%29%29%5C%5C%26%3D%5Cleft%5Bf_%2A%28X%29%5Cright%5D%5Cphi%5C%5C%26%3DX%28%5Cphi%5Ccirc+f%29%5C%5C%26%3D%5Cleft%5Bd%28%5Cphi%5Ccirc+f%29%5Cright%5D%28X%29%5C%5C%26%3D%5Cleft%5Bd%28f%5E%2A%5Cphi%29%5Cright%5D%28X%29%5Cend%7Baligned%7D&bg=e6e6e6&fg=333333&s=0 "\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 -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.
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“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