# Why is the pullback completely determined by $d f^\ast = f^\ast d$ in de Rham cohomology?

Fix a smooth map $f : \mathbb{R}^m \rightarrow \mathbb{R}^n$. Clearly this induces a pullback $f^\ast : C^\infty(\mathbb{R}^n) \rightarrow C^\infty(\mathbb{R}^m)$. Since $C^\infty(\mathbb{R}^n) = \Omega^0(\mathbb{R}^n)$ (the space of zero-forms) by definition, we consider this as a map $f^\ast : \Omega^0(\mathbb{R}^n) \rightarrow \Omega^0(\mathbb{R}^m)$. We want to extend $f^\ast$ to the rest of the de Rham complex in such a way that $d f^\ast = f^\ast d$.

Bott and Tu claim (Section I.2, right before Prop 2.1), without elaboration, that this is enough to determine $f^\ast$ . I can see why this forces e.g.

$\displaystyle\sum_{i=1}^n f^\ast \left[ \frac{\partial g}{\partial y_i} d y_i \right] = \sum_{i=1}^n f^* \left[ \frac{\partial g}{\partial y_i}\right] d(y_i \circ f)$,

but I don't see why this forces each term of the LHS to agree with each term of the RHS -- it's not like you can just pick some $g$ where $\partial g/\partial y_i$ is some given function and the other partials are zero.

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I claim that the function $y_j$ has the property that $\partial y_j / \partial y_i$ is the constant $1$ if $i = j$ and is the constant $0$ otherwise. –  Zhen Lin Dec 28 '11 at 15:09
Right, but you can't get a function with the property that $\partial g / \partial y_i$ is an arbitrary smooth function and the other partials are zero, because this forces $\partial(\partial g/\partial y_i)/\partial y_j = \partial^2 g / \partial y_i \partial y_j = \partial(\partial g / \partial y_j)/\partial y_i = 0$ for all $i \neq j$. (I was trying to do this without the assumption that $f^\ast$ preserved multiplication, which is apparently a mistake.) –  Daniel McLaury Dec 28 '11 at 15:47

$\newcommand\RR{\mathbb{R}}$I don't have the book here, but it seems you are asking why there is a unique extension of $f^*:\Omega^0(\RR^n)\to\Omega^0(\RR^m)$ to an appropriate $\overline f^*:\Omega^\bullet(\RR^n)\to\Omega^\bullet(\RR^m)$ such that $f^*d=df^*$. Here appropriate should probably mean that the map $\overline f^*$ be a morphism of graded algebras.

Now notice that the since $f^*$ is fixed on $\Omega^0(\RR^n)$ and the commutation relation with $d$ tells us that it is also fixed on the subspace $d(\Omega^0(\RR^n))\subseteq\Omega^1(\RR^n)$. The uniqueness follows from the fact that the subspace $\Omega^0(\RR^n)\oplus d(\Omega^0(\RR^n))$ of $\Omega^\bullet(\RR^n)$ generates the latter as an algebra.

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Dear Mariano: You wrote "morphism of graded algebras", but I think you meant "morphism of differential graded algebras". –  Pierre-Yves Gaillard Dec 28 '11 at 15:36
Actually, I separated the 'differential' part because it was singled out in the question itself, leaving for 'appropriate' simply the graded-algebra part :) Of course, the resulting map will be a morphism of differential graded algebras. –  Mariano Suárez-Alvarez Dec 28 '11 at 15:37
Ah, so $f^\ast$ should not only be $\mathbb{R}$-linear, but should preserve multiplication as well -- that's what I was missing. For some reason it didn't occur to me that the original pullback function preserved multiplication, so I never considered that I should want such a property for the extension. –  Daniel McLaury Dec 28 '11 at 15:41
Consider $h = g \circ Q_{i}$ where $Q_{i}$ sends $(a_1,...,a_n)$ to $(0,\dotsc, a_i, \dotsc, 0)$. Apply your observation to $h$.