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 $p$. $\qquad \clubsuit$

  • $\begingroup$ You've only shown that $\Omega^p(U)\to\Omega^p(V)$ is surjective. What makes you think you've done it globally? $\endgroup$ Aug 19, 2019 at 19:20
  • $\begingroup$ 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? $\endgroup$
    – peabody
    Aug 19, 2019 at 21:13
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    $\begingroup$ 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. $\endgroup$ Aug 19, 2019 at 21:38
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    $\begingroup$ Oh, by that I mean a closed embedding .... $\endgroup$ Aug 19, 2019 at 23:41
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    $\begingroup$ 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. $\endgroup$ Aug 20, 2019 at 16:33

1 Answer 1


This is just here so that I can mark the question as answered for people looking for it. The proof provided above should no longer have any errors, but if someone finds a better way of doing things (or an error), please post a new response, and I'll mark that as the preferred answer over this one. Thank you!


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