# Prove that the Pullback of a proper embedding is surjective on p-forms

This is from an old qualifying exam, and I want to check the answer that I have given. The question is

Let $$N$$ be a compact embedded submanifold of a manifold $$M$$. Show that $$\Omega^p(M) \rightarrow \Omega^p(N)$$ is surjective for all $$p$$.

Here $$\Omega^p(M)$$ is the vector space of smooth $$p$$-forms. Here is my attempt:

Let $$\iota: N \rightarrow M$$ be the inclusion map and pick a point $$q \in N$$. Let $$(U_q, (x^i))$$ be an open chart centered at $$\iota(q) = q$$ such that $$U_q\cap N = V_q$$ is a local $$k$$-slice of $$U_q$$. Then we know that $$(\iota^{-1}(V_q), (y^i)) = (V_q, (y^i))$$ where $$y^i(a) = x^i(a)$$ for $$a\in V_q$$ and $$i \in \{1, \ldots, k\}$$. Then for any $$\omega \in \Omega^p(N)$$, we can express $$\omega$$ in terms of these local coordinates as a sum over increasing multi-indices $$I$$ by $$\omega = {\sum_I}^\prime \omega_I dy^{i_1}\wedge \cdots \wedge dy^{i_p}$$ where each $$\omega_I$$ is a smooth coordinate function defined on $$V_q$$. Well, if we keep the same indexing set and define the $$p$$-form $$\eta$$ on $$U_q$$ by $$\eta = {\sum_I}^\prime (\eta_I) dx^{i_1}\wedge \cdots \wedge dx^{i_p}$$ with $$\eta_I(\iota(a)) = \omega_I(a)$$, then \begin{align*} \iota^*\eta &= {\sum_I}^\prime (\eta_I\circ \iota) d(x^{i_1}\circ \iota)\wedge \cdots \wedge d(x^{i_p}\circ \iota)\\ &= {\sum_I}^\prime \omega_I dy^{i_1}\wedge \cdots \wedge dy^{i_p}\\ &= \omega \end{align*} hence we have that the map $$\Omega^p(U_q) \rightarrow \Omega^p(V_q)$$ is surjective for all $$p$$.

Now, we know that $$\{U_q\}_{q\in N}$$ forms an open cover of $$N$$ and admits a smooth partition of unity $$\{\varphi_q\}_{q\in N}$$ subordinate to this open cover. Define $$\psi_q = \varphi_q\vert_{N}$$. Then the collection $$\{\psi_q\}_{q\in N}$$ will be a smooth partition of unity of $$N$$ subordinate to the open cover $$\{V_q\}_{q\in N}$$. Let $$\widetilde{\omega}\in \Omega^p(N)$$ be any smooth $$p$$ form on $$N$$. Since this form is smooth, we know that there is an indexed collection of $$p$$-forms $$\{\omega^q\}_{q\in N}$$ such that

$$\widetilde{\omega} = \sum_{q\in N}\psi_q\omega^q.$$

Now define the $$p$$-form $$\widetilde{\eta}\in \Omega^p(M)$$ by

$$\widetilde{\eta} = \sum_{q\in N}\varphi_q\eta^q$$

where each $$\eta^q$$ is such that $$\iota^*\eta^q = \omega^q$$ as given in the first portion of this proof and $$\widetilde{\eta}|_y = 0$$ for any $$y\not\in\bigcup_{q\in N}U_q$$. This will be defined and smooth on all of $$M$$ since $$supp(\varphi_q\eta^q)\subseteq U_q$$. Furthermore, since this partition of unity is locally finite, for any $$x \in N$$, we have that $$\widetilde{\omega}|_x = \iota^*\widetilde{\eta}|_x$$, and so $$\widetilde{\omega} = \iota^*\widetilde{\eta}$$. Hence, $$\Omega^p(M) \rightarrow \Omega^p(N)$$ is surjective for all $$p$$. $$\qquad \clubsuit$$

• You've only shown that $\Omega^p(U)\to\Omega^p(V)$ is surjective. What makes you think you've done it globally? Aug 19, 2019 at 19:20
• Leo, thank you for that example, that definitely shows that I have missed something. And @TedShifrin, thank you for always being so helpful. I see what you mean, and I have edited what I wrote to reflect what you said, but I'm still a bit stuck. Ideally, I would like to say that any $\widetilde{\omega} \in \Omega^p(N)$ restricts down to a $\omega\in\Omega(V_i)$ where finitely many $V_i$ cover $N$, but I'm having trouble seeing how this helps me since I can't really say how this would relate to some $p$-form $\widetilde{\eta}\in \Omega^p(\bigcup_1^n U_i)$, can I? Aug 19, 2019 at 21:13
• You were right that you need to glue local forms together with a partition of unity. You actually don't need compactness, I think,, just that you have an embedded submanifold. Aug 19, 2019 at 21:38
• Oh, by that I mean a closed embedding .... Aug 19, 2019 at 23:41
• Local finiteness tells you that the sum makes sense (pointwise). I don't think you need to do the intersections. Strictly speaking, though, why is $\tilde\eta$ defined on all of $M$? You need to make some remark(s) to justify this. Aug 20, 2019 at 16:33