formula for f*$\omega$ I have the following problem:

Let $U,V\subset\mathbb{R}^n$ be open subsets, $f:U \rightarrow V$ a diffeomorphism and $\omega \in \Omega^nV$. Find a formula for $f^*\omega$.

My problem is that I don't know the meaning of $f^*\omega$.
$\Omega^nV$ is a vector space of k-forms, which are continuously differentiable once.
 A: Well, $\omega$ is something like a sum of terms of the form
$$g_1(y_1,y_2, \dots y_n)\wedge g_2(y_1,y_2, \dots y_n)\wedge \dots \wedge g_k(y_1,y_2, \dots y_n)$$
and given 
$$ f = (f_1(x_1,x_2, \dots x_n), f_2(x_1,x_2, \dots x_n), \dots f_n(x_1,x_2, \dots x_n))
$$
you just have $f^* \omega$ as a sum of terms of the form
$$g_1(f_1(x_1,x_2, \dots x_n), f_2(x_1,x_2, \dots x_n), \dots f_n(x_1,x_2, \dots x_n))\wedge g_2(f_1(x_1,x_2, \dots x_n), f_2(x_1,x_2, \dots x_n), \dots f_n(x_1,x_2, \dots x_n))\wedge \dots \wedge g_k(f_1(x_1,x_2, \dots x_n), f_2(x_1,x_2, \dots x_n), \dots f_n(x_1,x_2, \dots x_n)).$$
That is, just substitute $\bf y$ in the image with $\bf f (\bf x)$ 
A: $f^*\omega$ is the pull-back of $\omega$. If $f: U \to V$ and $\omega$ is a differential form on $V$ then $f^*\omega$ is a differential form on $U$. How do we define this differential form on $U$? Given tangent vectors $u_1,\ldots,u_k$ we define $(f^*\omega)(u_1,\ldots,u_k)=\omega(f_*u_1,\ldots,f_*u_k)$, where $f_* : TU \to TV$ is the differential of $f$. The differential is a linear map whose matrix representation is the Jacobian matrix.
Calculate the Jacobian matrix. Apply the Jacobian matrix to each of the $u_i$ to give $f_*u_i$. Then apply the differential form $\omega$ on the $f_*u_i$.
