Volume and orientation application -Calculus of manifolds If $\omega$ is the volumen element of $V$ determined by $T$ and $\mu$, and $f\colon \mathbb{R}^n\to V$ is an isomorphism such that $f^*T=\langle,\rangle$, and such that $[f(e_1),\ldots,f(e_n)]=\mu$, show that $f^*\omega=det$
I am trying to do exercise 4-4 of calculus of manifolds Spivak, The hypothesis of the exercise is very similar to Theorem 4-2.(Calculus of manifolds - Spivak)

what I have tried to give an ortnormal base $[f(e_1),\ldots,f(e_n)]$, such that the pullback $f^*T(e_i,e_j)=\delta_{ij}$, and $f^*\omega(e_1,\ldots,e_n)=\omega(e_1,\ldots,e_n)$.
They are only ideas, since I have not been able to solve it, any help or suggestion is well received
 A: The thorem you attached isn't really what you should be looking at. This exercise amounts to reading carefully the pages on how an inner product and orientation determines a volume element.
Let us briefly recall how this goes. First a definition: if $W$ is an $n$-dimensional real vector space then a volume element is by definition a non-zero alternating $(0,n)$ tensor  over $W$; i.e an element $\eta \in \mathcal{A}^n(W)\setminus\{0\}$ (recall that this space is $1$-dimensional so we're essentially choosing a basis for this space).
On an arbitrary vector space $W$ we don't have a specific way of deciding which volume element to take: if we choose one, then any non-zero scalar multiple will do. However, if we have an oriented inner-product space $(W,g,\nu)$ then there is a unique choice, because there is a unique $\eta \in \mathcal{A}^n(W)\setminus\{0\}$ such that for some (and hence any) ordered orthonormal basis $\{w_1,\dots, w_n\}$ of $W$ with the correct orientation $[w_1,\dots, w_n]=\nu$, we have
\begin{align}
\eta(w_1,\dots, w_n)&=+1.
\end{align}

I intentionally avoided the notation $V$ in this discussion because for your question we'll be taking $W=\Bbb{R}^n$, $g=\langle \cdot,\cdot\rangle$ the standard inner product, and the usual orientation $\nu=[e_1,\dots, e_n]$.
For your question, you're given an isomorphism $f$ along with a volume element $\omega$ on $V$. You should know (pretty much by basic definition unwinding) that $f^*\omega$ is an alternating $(0,n)$ tensor on $\Bbb{R}^n$, i.e $f^*\omega\in \mathcal{A}^n(\Bbb{R}^n)$. But of course we already know that $\det$ spans this $1$-dimensional vector space (because as mentioned in the comments this is usually how one defines the determinant... or if you take a more abstract definition of the determinant this also follows easily). Hence, there is a $c\in\Bbb{R}$ such that
\begin{align}
f^*\omega &= c\cdot \det
\end{align}
Ideally, we'd like $c=1$, so how do we show that? Simple. Take the standard basis $\{e_1,\dots, e_n\}$ of $\Bbb{R}^n$ then we have
\begin{align}
c&= c\cdot 1\\
&= c\cdot \det(e_1,\dots, e_n)\\
&=(f^*\omega)(e_1,\dots, e_n)\\
&= \omega(f(e_1),\dots, f(e_n))
\end{align}
Now, based on everything I've mentioned above can you justify why this last line equals $1$? (Of course you need to use your assumptions on $f$ at this stage).
