Computing the pull-back with exterior derivative I don't understand how to do the following question:

Property 5 is $\phi$*d$\omega$=d$\phi$ *$\omega$
Thank you!
 A: You have 
$$
F : \Bbb R^3 \to \Bbb R^6: (x, y, z) \mapsto (x, y, z, xyz, x^3 - y, 6x + 3z)
$$
And you have a $2$-form, $\omega$ on $\Bbb R^6$. You'd like to "compute" 
$$
\eta = F^{*}\omega,
$$
whatever that might mean, which should be a $2$-form on $\Bbb R^3$. That means that $\eta$ should be something that looks like
$$
\eta(p, q, r) = f(p,q,r) dp \wedge dq + g(p,q,r) dq \wedge dr  + h(p,q,r) dp \wedge dr
$$
i.e., you're supposed to write down the functions $f, g, h$. If you prefer a notation with lots of superscripts, etc., then you want
$$
\eta(x^1, x^2, x^3) = f(x^1, x^2, x^3)  dx^1 \wedge dx^2 + \ldots,
$$
but frankly, I think that's worse to read. To each their own. 
I have no idea what the hint is getting at, but let's look at what 
$$
\eta(p,q,r) (e_1, e_2)
$$
should be. On the one hand, it should be 
\begin{align}
\eta(p, q, r)(e_1, e_2)
&= f(p,q,r) dp \wedge dq (e_1, e_2)+ g(p,q,r) dq \wedge dr (e_1, e_2) + h(p,q,r) dp \wedge dr(e_1, e_2)\\
&= f(p,q,r) dp \wedge dq (e_1, e_2)+ g(p,q,r)\cdot 0  + h(p,q,r) \cdot 0\\
&= f(p,q,r) \cdot 1
\end{align}
Hold that thought. 
On the other hand, it should also be 
\begin{align}
\eta(p, q, r)(e_1, e_2) 
&= (F^* \omega)(p,q,r) (e_1, e_2)\\
&= \omega(F(p,q,r))(F_* (p,q,r) (e_1), F_* (p,q,r) (e_2))\\
&= \omega((p, q, r, pqr, p^3 - r, 6p + 3r))(F_* (p,q,r) (e_1), F_* (p,q,r) (e_2))\\
&= ( (p+r)~dw^3 \wedge dw^4 + 2 ~dw^6 \wedge dw^5 + q ~dw^1 \wedge dw^2)(F_* (p,q,r) (e_1), F_* (p,q,r) (e_2))\\
\end{align}
Now we need to compute $F_{*}$, which is just the derivative of $F$, so at $(p,q,r)$, it's multiplication by the matrix 
$$
\pmatrix{1 &  0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
qr & pr & pq \\
3p^2& -1 & 0 \\
6 & 0 & 3}
$$
Multiplying this by $e_1$ gives the first column; multiplying by $e_2$ gives the second column. So 
\begin{align}
\eta(p, q, r)(e_1, e_2) 
&= ( (p+r)~dw^3 \wedge dw^4 + 2 ~dw^6 \wedge dw^5 + q ~dw^1 \wedge dw^2)(F_* (p,q,r) (e_1), F_* (p,q,r) (e_2))\\
&= ( (p+r)~dw^3 \wedge dw^4 + 2 ~dw^6 \wedge dw^5 + q ~dw^1 \wedge dw^2)(\pmatrix{1 \\
0 \\
0 \\
qr\\
3p^2 \\
6 }, \pmatrix{  0  \\
 1  \\
 0  \\
 pr  \\
 -1  \\
 0 })\\
\end{align}
So all we have to do is evaluate $dw^3 \wedge dw^4$ on that pair of vectors (we get $0$), $dw^6 \wedge dw^5$ (messier), and $dw^1 \wedge dw^2$. This last one gives $1 \cdot 1 - 0 \cdot 0 = 1$. So the middle one remains. Calling the two vectors $v_1$ and $v_2$, we need to compute
\begin{align}
s &= dw^6(v_1) dw^5(v_2) - dw^6(v2) dw^5(v_1) \\
&= 6 \cdot (-1) - 0 \cdot 3p^2 \\
&= -6.
\end{align}
So we end up with 
\begin{align}
\eta(p, q, r)(e_1, e_2) 
&= ( (p+r)~dw^3 \wedge dw^4 + 2 ~dw^6 \wedge dw^5 + q ~dw^1 \wedge dw^2)(F_* (p,q,r) (e_1), F_* (p,q,r) (e_2))\\
&= ( (p+r)~dw^3 \wedge dw^4 + 2 ~dw^6 \wedge dw^5 + q ~dw^1 \wedge dw^2)(\pmatrix{1 \\
0 \\
0 \\
qr\\
3p^2 \\
6 }, \pmatrix{  0  \\
 1  \\
 0  \\
 pr  \\
 -1  \\
 0 })\\
&= ( (p+r)~\cdot 0  + 2 \cdot (-6) + q \cdot 1\\
&= -12 + q
\end{align}
from which we conclude that $f(p, q, r) = -12 + q$.
Now all you have to do is repeat that kind of computation to find the functions $g$ and $h$, and you're done. 
