# $(f\circ h)^* \omega = h^*f^*\omega$ - Legit now?

Three pullback identities are required to prove on Guillemin and Pollack's Differential Topology on Page 164. I am not sure if I get it since it is the first time I play with them.

I hope I got the first two right, but I was not able to proceed the third, because I don't see the commutativity of the pullback $df_x^*$ and the form $\omega$.

Definition. If $f: X \to Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, define a $p$-form $f^*\omega$ on $X$ as follows: $$f^*\omega(x) = (df_x)^*\omega[f(x)].$$

$f^*(w_1 + w_2) = f^*w_1 + f^* w_2$

$f^*(w \wedge \theta) = (f^*w) \wedge (f^* \theta)$

$\mathbf{(f\circ h)^* \omega = h^*f^*\omega}$

Following the definition, the RHS is: \begin{eqnarray*} d(f \circ h)_x^* \omega [(f \circ h)(x)]& =&\omega( d(f \circ h)_x(x))\\ & =&\omega( df_{h(x)} \circ dh_x)(x))\\ & =&\omega( df_{h(x)} dh_x(x))\\ & =&df_{h(x)}^*\omega( dh_x(x))\\ & =&f^*\omega( dh_x(x))\\ & =&h^*f^*\omega( x)\\ \end{eqnarray*}

Let $V_1, \dots, V_p \in T_x X$ be a collection of $p$ tangent vectors to $X$ at $x$. Then we compute that for any $x \in X$, \begin{align} ((f \circ h)^\ast \omega)_x(V_1, \dots, V_p) & = \omega_{f(h(x))}(d(f \circ h)_x V_1, \dots, d(f \circ h)_x V_p) \\ & = \omega_{f(h(x))}(df_{h(x)} dh_x V_1, \dots, df_{h(x)} dh_x V_p) \\ & = (f^\ast \omega)_{h(x)} (dh_x V_1, \dots, dh_x V_p) \\ & = (h^\ast f^\ast \omega)_x(V_1, \dots, V_p), \end{align} where we used the chain rule in going from the first line to the second line, and in the last two lines we just undid the definition of the pullback twice. Hence $$(f \circ h)^\ast \omega = h^\ast f^\ast \omega.$$
• Thank you Henry! I pondered for a while, and here's a point I'm not quite clear about - at the very last step, why you bring $h$ out, rather than seeing $\omega(dh_x(x))$ as a whole and get $h^*\omega$ as I did? I'd like to know clearly where I went wrong. – WishingFish Aug 2 '13 at 5:35
• Some parts of your notation don't make sense -- $dh_x$ is a map between tangent spaces, so it should act on tangent vectors, not $x$. So $dh_x(x)$ doesn't make sense. In any case, bringing $h$ out" is the same thing as "seeing $\eta_{h(x)}(dh_x V_1, \dots, dh_x V_p)$ as a whole," where $\eta = f^\ast \omega$. I just use the definition of the pullback here: $\eta_{h(x)}(dh_x V_1, \dots, dh_x V_p) = (h^\ast \eta)_x(V_1, \dots, V_p) = (h^\ast f^\ast \omega)_x(V_1, \dots, V_p)$. – Henry T. Horton Aug 2 '13 at 5:54
• The definition was given at Guillemin and Pollack's Differential Topology. I think the author intended $x$ to be a vector in the tangent space....? – WishingFish Aug 2 '13 at 6:17
• I see your quoted definition in Guillemin and Pollack, but I don't see them write $dh_x(x)$ anywhere. They're always taking $x$ to be a point in the manifold, not a tangent vector. – Henry T. Horton Aug 2 '13 at 6:39
• Oh! I see why I was wrong, thanks! Though, what is $V_1, \dots, V_p$? I don't quite get what is "a collection of vector field". Thank u. – WishingFish Aug 2 '13 at 6:59