# Pullback of a differential form by a local diffeomorphism

Suppose I have to smooth oriented manifolds, $M$ and $N$ and a local diffeomorphism $f : M \rightarrow N$. Let $\omega$ be a differential form of maximum degree on $N$, let's say, $r$. How can I rewrite $$\int_N \omega$$ in terms of the integral of the pullback of $\omega$ by $f$, $f^*\omega$?

So, I know that when $f$ is a diffeomorphism, then $$\int_N\omega = \pm \int_M f^*\omega$$ depending on whether $f$ preserves the orientation or not. But that's in part due to the fact that $f$ is bijective, but that condition is removed when assuming $f$ is a local diffeomorphism. So is there a nice way to write that integral in terms of the pullback?

Yeah, and you can make it work even if $$f$$ is not a local diffeomorphism, only a proper map. The relevant notion is that of the degree of a smooth proper map between oriented manifolds of the same dimension.

Assume for simplicity that $$M,N$$ are closed (compact, without boundary), non-empty and connected and let $$f \colon M \rightarrow N$$ be an arbitrary smooth map. Since $$M,N$$ are oriented, we can a generator $$[\omega_1] \in H^{\text{top}}(M)$$ such that $$\omega_1$$ is consistent with the orientation on $$M$$ and $$\int_M \omega_1 = 1$$. Choose $$[\omega_2]$$ similarly for $$N$$. The map $$f$$ induces a map $$f^{*} \colon H^{\text{top}}(N) \rightarrow H^{\text{top}}(M)$$ on cohomology and since the top cohomology groups are one-dimensional, we must have $$f^{*}([\omega_2]) = c [\omega_1]$$ for some $$c \in \mathbb{R}$$. This $$c = \deg(f)$$ is called the degree of $$f$$ and is a priori a real number. However, it can be shown that $$c$$ is in fact an integer which can be computed by counting the number of preimages of a regular value $$p \in N$$ with appropriate signs which take the orientations of $$M,N$$ into consideration.

Knowing that, given $$\omega \in \Omega^{\text{top}}(N)$$, write $$[\omega] = c[\omega_2]$$ for some $$c \in \mathbb{R}$$. Then

$$[f^{*}(\omega)] = f^{*}([\omega]) = f^{*}(c[\omega_2]) = c \deg(f) [\omega_1]$$

so

$$\int_M f^{*}(\omega) = \deg(f) c \int_N \omega_1 = \deg(f) c = \deg(f) c\int_N \omega_2 = \deg(f) \int_N \omega.$$

In particular, if $$f$$ is a diffeomorphism, $$\deg(f) = \pm 1$$ so this generalizes your starting point. If $$f$$ is a local diffeomorphism then $$f \colon M \rightarrow N$$ is a covering map and $$\deg(f)$$ will be the number of points in an arbirary fiber, counted with appropriate signs. For much more details and proofs, see the book "Differential Forms in Algebraic Topology".

• For your particular case, you can avoid mentioning cohomology by showing first (assuming $M,N$ are compact) that $f$ is a covering map. Then, over a basic neighborhood $U \subseteq N$, we have $f^{-1}(U) = \sqcup {U_i}$ where $f|_{U_i} \colon U_i \rightarrow U$ is a diffeomorphism which reverses or preserves orientation. Hence $\int_{f^{-1}(U)} f^{*}(\omega) = \sum \pm \int_{N} \omega$. Then you can get the global result by a partition of unity after showing that "the signs stay constant" on the intersections so there isn't any unwanted cancellations. – levap Jan 25 '17 at 21:19