Pullback of a form under the retraction $r\colon \mathbb{R}^n\setminus\{0\}\to S^{n-1}$. The following is from Spivak's DG Lemma 7 in Chapter 8, but I'm muddled in a computation.
Define two $(n-1)$-forms on $\mathbb{R}^n\setminus\{0\}$ by
$$
\sigma=\sum_{i=1}^n(-1)^{i-1}x^idx^1\wedge\cdots\wedge\widehat{dx^i}\wedge\cdots\wedge dx^n
$$
$$
\omega=\frac{1}{|x|^n}\sigma.
$$
If $r(x)=\frac{x}{|x|}$ is the retraction, and $\iota\colon S^{n-1}\to\mathbb{R}^n\setminus\{0\}$ the inclusion, then $\omega=r^*\iota^*\sigma$. 
Pick $p\in\mathbb{R}^n\setminus\{0\}$, and let $v_1,\dots,v_{n-1}$ be tangent vectors, then
$$
\begin{align*}
r^*\iota^*\sigma_p(v_1,\dots,v_{n-1})&=(\iota\circ r)^*\sigma_p(v_1,\dots,v_{n-1})\\
&=\sigma_{\iota r(p)}(d(\iota\circ r)_p(v_1),\dots,d(\iota\circ r)_p(v_{n-1}))\\
&=\sigma_{p/|p|}(d(\iota\circ r)_p(v_1),\dots,d(\iota\circ r)_p(v_{n-1}))\\
&=\left(\sum_{i=1}^n(-1)^{i-1}x^i(p/|p|)dx^1|_{p/|p|}\wedge\cdots\wedge\widehat{dx^i|_{p/|p|}}\wedge\cdots\wedge dx^n|_{p/|p|}\right)(d(i\circ r)_p v_1,\dots,d(\iota\circ r)_p v_{n-1})
\end{align*}
$$
On the other hand, 
$$
\begin{align*}
\omega_p(v_1,\dots,v_{n-1})&=\frac{1}{|p|^n}\sigma_p(v_1,\dots,v_{n-1})\\
&=\frac{1}{|p|^n}\left(\sum_{i=1}^n(-1)^{i-1}x^i(p)dx^1|_p\wedge\cdots\wedge\widehat{dx^i|_p}\wedge\cdots\wedge dx^n|_p\right)(v_1,\dots,v_{n-1})\\
\end{align*}
$$
How can I see that these two expressions are equal to each other? Thanks.
 A: To simplify, denote $$X=\sum_{i=1}^nx^i\frac{\partial}{\partial x^i}$$ and $\Omega=dx^1\wedge dx^2 \cdots \wedge dx^n$, then $\sigma=i(X)\Omega$, where $i$ is the Interior Product operator, that is, $\sigma(\cdots)=\Omega(X,\cdots)$.  $\forall p\in \mathbb{R}^n$, let $q=r(p)$. Note that $X$ has the following property: $$X_q=\sum_{i=1}^n\frac{x^i(p)}{|p|}\frac{\partial}{\partial x^i}=\frac{X_p}{|p|}$$
Now let $f=\iota\circ r$, to compute $f^*\sigma$ at $p$, let's choose an orthonormal basis at $p$ as $$\{e_1=\frac{p}{|p|}, e_2, \cdots, e_n\}$$
It's easy to verify the following for the differential map $f_*$:
$$f_*(e_1)=0,\ f_*(e_i)=\frac{e_i}{|p|}, \forall i=2,\cdots,n$$
So we need only to compute $f^*\sigma$ against $\{e_2,\cdots, e_n\}$, and $$(f^*\sigma)_p(e_2,\cdots,e_n)=\sigma_q(f_*(e_2),\cdots,f_*(e_n))$$
$$=\Omega_q(\frac{X_p}{|p|},\frac{e_2}{|p|},\cdots,\frac{e_n}{|p|})$$
$$=\frac{1}{|p|^n}\Omega_p(X_p,e_2,\cdots,e_n)$$
$$=\frac{1}{|p|^n}\sigma_p(e_2,\cdots,e_n)$$
That is $$f^*\sigma=\frac{1}{|p|^n}\sigma$$
