# form exact $\Leftrightarrow$ pull-back exact

Is the form exact $\Leftrightarrow$ pull-back exact? Since $$f^*\omega = \omega \circ df,$$ which seems irrelavant. Because the composition with $df$ does not change $\omega$ is exact or not.

The definition I was trying to use:

Suppose $A: V \to W$ is a linear map. Then the transpose map $A^*: W^* \to V^*$ extends to the exterior algebras, $A^*: \Lambda^p(W^*) \to \Lambda^p(V^*)$ for all $p>0$. If $T \in \Lambda^p(W^*)$, just define $A^* T \in \Lambda^p(V^*)$ by $$A^*T(v_1, \dots, v_p) = T(Av_1, \dots, Av_p)$$ for all vectors $v_1, \dots, v_p \in V$. (Page 159)

• You have $d(f^\ast\alpha) = f^\ast d\alpha$. – Daniel Fischer Aug 9 '13 at 20:39
• You should take another look at the definition of the pullback of a differential form, because the formula you wrote down isn't exactly true. In any event, as Daniel Fischer points out, $d(f^\ast \omega) = f^\ast d\omega$, which answers your question affirmatively. – Branimir Ćaćić Aug 9 '13 at 20:40
• Thanks @DanielFischer , so I got a really dull question here: I got the part that the pull-back of an exact form is exact, since $f^* d\alpha = d(f^* \alpha).$ Hence by contrapositive, the pull back of $\beta$ is not exact, so $\beta$ is not exact. But how can I show the other way? – WishingFish Aug 9 '13 at 21:04
• The other way doesn't generally hold. The pull-back $f^\ast\omega$ can be exact without $\omega$ being exact. For example if $f \colon X \to Y$ where $\dim X < \dim Y$, then the pull-back of every $\dim Y$-form is exact (it's $0$), regardless of whether the pulled-back form is exact. Or if $f$ is constant, all pull-backs (except for $0$-forms) are $0$, hence exact. If $f$ is a diffeomorphism, the pull-back by $f^{-1}$ is the inverse of the pull-back by $f$, and that makes the equivalence true. You need some pretty strict conditions on $f$ for the equivalence to hold. – Daniel Fischer Aug 9 '13 at 21:13
• What an excellent answer, @DanielFischer. Thanks! – WishingFish Aug 9 '13 at 23:01