In Loring Tu's book, An Introduction to manifolds, he uses the following proposition to prove that the pullback of a smooth $k$ form is a smooth $k$ form:

Proposition: Let $F:N\rightarrow M$ be a smooth map. If $\omega$ is a smooth k-form then $$\mathrm dF^{*}\omega=F^{*}\mathrm d\omega.$$

However, my question is: Isn't the exterior derivative, $\mathrm d$ is a map $\Omega^{k}(M)\rightarrow \Omega^{k+1}(M)$ (space of smooth $k$, $k+1$ forms), so isn't he assuming that $F^{*}\omega$ is smooth?

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    $\begingroup$ There's no need to assume. If $F$ and $\omega$ are smooth, then $F^*\omega$ is automatically smooth as well. $\endgroup$
    – Kajelad
    Commented Aug 10, 2020 at 16:44
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    $\begingroup$ Yes. One can show that the pullback of a smooth form by a smooth function is smooth. $\endgroup$
    – Kajelad
    Commented Aug 10, 2020 at 16:54
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    $\begingroup$ Maybe what you've thinking of is that you can use the fact that the pullback of any smooth form by $F$ is a smooth form and the fact that $dF^*\omega=F^*d\omega$ to construct an explicit formula for the pullback of the smooth $k$-form? $\endgroup$ Commented Aug 10, 2020 at 18:14
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    $\begingroup$ @RachidAtmai: I do not know how to define $d\phi$ unless $\phi$ is at least a $C^1$ differential form. We're not going to talk about currents and distributional derivatives here. Smoothness of $F^*\omega$ is a separate question from computing exterior derivatives, as I already said. When we take exterior derivative, only certain combinations of partial derivatives appear, and that does not establish smoothness. $\endgroup$ Commented Aug 10, 2020 at 22:08
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    $\begingroup$ I do not own the book, so I can't verify whether OP is misinterpreting. But I surmise he is. The proof that the pullback of a smooth form by a smooth function is an entirely self-contained, separate argument. Start by showing $f^*dx^i$ is smooth. $\endgroup$ Commented Aug 10, 2020 at 22:11

2 Answers 2


The way it stands, the placement of Proposition 19.5 is a mistake, because $F^*\omega$ needs to be $C^{\infty}$ before one can take its exterior derivative. To fix this, in Proposition 19.7, replace the justification "(Proposition 19.5)" by "(Proposition 17.10)," and then move Proposition 19.7 to before Proposition 19.5.

I see that Arctic Char has proposed the same solution a while ago. I give it my ringing endorsement.

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    $\begingroup$ Hi @Loring! :) :) $\endgroup$ Commented Aug 15, 2020 at 6:13

After reading the corresponding section I agree that the author is wrong in claiming that $\mathrm d F^*\omega = F^* \mathrm d\omega$ is used to show that $F^* \omega$ is smooth. This is not necessary.

First recall (I am using the second edition):

Proposition 19.7: If $F : N \to M$ is a $C^\infty$ map of manifolds and $\omega$ is a $C^\infty$ $k$-form on $M$, then $F^*\omega$ is a $C^\infty$ $k$-form on $N$.

Proof (Sketch) given in the book: In a local coordinates, $$\omega = \sum_I a_I \mathrm dy^{i_1} \wedge \cdots \wedge \mathrm dy^{i_k}.$$ for some local smooth functions $a_I$. Then

\begin{align} F^*\omega &= \sum (F^* a_I) F^* \mathrm dy^{i_1} \wedge \cdots \wedge F^* \mathrm dy^{i_k} \\ &= \sum (F^* a_I) \mathrm d(F^* y^{i_1}) \wedge \cdots \wedge \mathrm d (F^* y^{i_k}) \ \ \ \ \ (\text{Proposition }19.5)\\ &= \cdots \\ &= \sum (a_I \circ F)\frac{\partial (F^{i_1}, \cdots, F^{i_k})}{\partial (x^{j_1} \cdots x^{j_k})} \mathrm dx^J. \end{align}


$$(a_I \circ F)\frac{\partial (F^{i_1}, \cdots, F^{i_k})}{\partial (x^{j_1} \cdots x^{j_k})}$$ are smooth, the author concludes that $F^*\omega$ is smooth.

Proposition 19.5 says that for any smooth $k$-form $\omega$ we have $F^* \mathrm d \omega = \mathrm d F^* \omega$.

As already pointed out by TedShifrin in the comment, only $F^* dy^{i_l} = d (F^* y^{i_l})$ is needed to show Proposition 19.7, and the proof in the book is exactly using just that. This fact is proved in the previous section (Proposition 17.10).

So I think it might be a typo to use Proposition 19.5 to prove Proposition 19.7. Indeed he needs only to use 17.10. Also it is confusing to put Proposition 19.5 before Proposition 19.7, that is, showing $F^* \mathrm d\omega = \mathrm d F^*\omega$ without first showing $F^*\omega$ is smooth. I did not check the whole book, but I guess the concept of $C^1$-differential form is not introduced. So it does not really make sense to talk about $\mathrm d F^*\omega$ without first showing that $F^*\omega$ is $C^\infty$, at least in the context of this book.

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    $\begingroup$ This clears everything up. Thank you. $\endgroup$
    – user643073
    Commented Aug 11, 2020 at 6:18
  • $\begingroup$ Arctic Char, nice job getting Loring Tu's 'ringing endorsement' $\endgroup$
    – BCLC
    Commented May 1, 2021 at 3:46

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