Pullback Calculation If we define the 2-form $\omega=\frac{1}{r^3}(x_1dx_2\wedge dx_3+x_2dx_3\wedge dx_1+x_3dx_1\wedge dx_2)$ with $r=\sqrt{x_1^2+x_2^2+x_3^2}$
If we now define $x(\theta,\phi)=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$ for $\theta\in(0,\pi)$ and $\phi\in (0,2\pi)$
I now want to show that the pullback of this map is:
$x^*\omega=\sin\theta d\theta\wedge d\phi$ 
Now my definition of the pullback of a map $c:[a,b]\rightarrow D$ is $\alpha(c'(t))dt$ but am unsure as to how to proceed? I can see that this curve $x$ is a parametrization of the sphere in $\mathbb{R}^3$ but after that I am a bit lost?
Thanks for any help
 A: A more general equation for how to compute pullbacks is as follows: If $f: M \to N$ is a smooth map of smooth manifolds and $\omega$ is an $n$-form on $N$, written in local coordinates as 
$$ \omega = \sum_I \omega_I dy^{i_1} \wedge \cdots \wedge dy^{i_n}$$
then $f^*\omega$ can be written in induced coordinates as
$$ f^*\omega = \sum_I (\omega_I \circ f) d(y^{i_1} \circ f) \wedge \cdots \wedge d(y^{i_n} \circ f)$$
where you think of $y^i$ as the function which picks out the $i^{th}$ coordinate of $f$.
Your $\omega$ is given by
$$ \omega = \omega_{23} dx_2 \wedge dx_3 + \omega_{31} dx_3 \wedge dx_1 + \omega_{12} dx_1 \wedge dx_2$$
where $\omega_{23} = \frac{x_1}{r^3}, \omega_{31} = \frac{x_2}{r^3}$ andd $\omega_{12} = \frac{x_3}{r^3}$. So now you just use the formula: I will do the $\omega_{23} dx_2 \wedge dx_3$ term as an example:
$$ f^*(\omega_{23} dx_2 \wedge dx_3) = (\omega_{23} \circ f) d(x_2 \circ f) \wedge d(x_3 \circ f) $$
Now 
\begin{align*}
\omega_{23} \circ f &= \frac{\sin\theta \cos\phi}{\sqrt{\sin^2\theta\cos^2\phi + \sin^2\theta \sin^2\phi + \cos^2\theta}^3} = \sin\theta\cos\phi \\
d(x_2\circ f) &= d(\sin\theta\sin\phi) = \cos\theta \sin\phi d\theta + \sin\theta \cos\phi d\phi \\
d(x_3\circ f) &= d(\cos\theta) = -\sin\theta d\theta.
\end{align*}
Putting it all together, we get
\begin{align*}
(\omega_{23} \circ f) d(x_2 \circ f) \wedge d(x_3 \circ f) &= (\sin\theta\cos\phi)(\cos\theta\sin\phi d\theta + \sin\theta\cos\phi d\phi) \wedge (-\sin\theta d\theta ) \\
&= \sin^3\theta\cos^2\phi d\theta \wedge d\phi
\end{align*}
(Of course, modulo mistakes in my computations!)
