Pushforward injective

Question

Let $f : M \rightarrow N$ be a smooth surjective map between smooth manifolds.

Now, consider a 2-form $\omega$ on $T_pN$. Does it now follow that the pullback satisfies? $f^* d \omega =0 \Rightarrow d \omega=0$

Answer 1

Yes. In fact, this is true for any $k$-form, regardless of whether or not it is in the image of $d$.

To see this, let $\omega$ be any smooth $k$-form on $N$. Let $U\subseteq N$ denote the open set consisting of all $p\in N$ for which $\omega_p$ is not identically $0$. We will assume $U\neq \emptyset$, and then prove that $f^\ast \omega$ is not the $0$ form.

Let $X\subseteq M$ denote the critical set of $f$, that is $$X = \{m\in M: d_m f \text{ does NOT have full rank}\}.$$

By Sard's theorem, $f(X)$ has measure $0$ in $N$. This implies that $N\setminus f(X)$ is dense, so $N\setminus f(X)$ intersects every open set. In particular, there is an $n\in (N\setminus f(X))\cap U$.

Since $f$ is surjective, there is an $m\in M$ with $f(m) = n$ and since $n\notin f(X)$, we may assume $d_m f$ has full rank. Now, since $\omega_n\neq 0$, there are $k$ tangent vectors $v_1,..., v_k\in T_n N$ with $\omega_n(v_1,...,v_k) \neq 0$.

Since $d_m f$ has full rank, it is surjective, so we can find $x_1,...,x_k\in T_m M$ with $d_m f(x_i) = v_i$.

In particular, \begin{align*} (f^\ast \omega)_m(x_1,...,x_k) &= \omega_{f(m)}(d_m f(x_1), ..., d_m f(x_k))\\ &= \omega_n(v_1,...,v_k)\\ &\neq 0.\end{align*}

Answer 2

By Sard's theorem, there is a dense subset $U \in N$ so that $Df$ is sujective on $f^{-1}(U)$. Then $d\omega$ is zero on $U$: this is a general fact in linear algebra:

Let $L : V\to W$ be a surjective linear map. If

$$L^*\alpha(X, Y, Z) = \alpha(LX, LY, LZ) = 0$$

for all $X, Y, Z\in V$, then

$$\alpha (a, b, c) = 0$$

for all $a, b, c \in W$, as $L$ is surjective. (Thus $\alpha = 0$)

Thus $d\omega$ is zero on $U$. As $U$ is dense, $d\omega =0$ on $N$.