I have a question to the proof of the following statement:
Let $e_1,\dotso, e_n$ be a basis of a real vector space $V$, let $\delta^1,\dotso, \delta^n$ be the dual basis and $1\leq\nu_1<\dotso<\nu_k\leq n$.
Then is
$\delta^{\mu_1}\wedge\dotso\wedge\delta^{\mu_k}(e_{\nu_1},\dotso, e_{\nu_k})=\begin{cases} sgn(\tau)\\ 0\end{cases}$,
with $\mu_i=\nu_{\tau(i)}$
The proof is simple, but there is one key thing, I do not understand, why it is necessary.
Proof:
Let $V_0$ be the k-dimensional subspace of $V$, which is generated by $e_{\nu_1},\dotso, e_{\nu_k}$, and $\iota:V_0\hookrightarrow V$ the inclusion. Since for every linear function $f$ we have the equality
$f^\ast\omega\wedge f^\ast\eta=f^\ast(\omega\wedge\eta)$
it is
$\delta^{\mu_1}\wedge\dotso\wedge\delta^{\mu_k}(e_{\nu_1},\dotso, e_{\nu_k})=\iota^\ast\delta^{\mu_1}\wedge\dotso\wedge\iota^\ast\delta^{\mu_k}(e_{\nu_1},\dotso, e_{\nu_k})$.
Is $(\mu_1,\dotso, \mu_k)$ not a permutation of $\nu_1<\dotso<\nu_k$, so is $i\neq j$ with $\mu_i=\mu_j$ and therefor $\delta^{\mu_i}\wedge\delta^{\mu_j}=0$, or it exists an $i$ with $\mu_i\neq\nu_j$ for every $j$. Then is $\iota^\ast\delta^{\mu_i}=0$.
If $(\mu_1,\dotso, \mu_k)$ is generated by a permutation $\tau$ from $(\nu_1,\dotso,\nu_k)$, hence $\mu_i=\nu_{\tau(i)}$, so is
$\iota^{\ast}\delta^{\mu_1}\wedge\dotso\wedge\iota\delta^{\mu_k}(e_{\nu_1},\dotso, e_{\nu_k})=sgn(\tau)\cdot\iota^\ast\delta^{\nu_1}\wedge\dotso\iota^{\ast}\delta^{\nu_k}(e_{\nu_1},\dotso, e_{\nu_k})=sgn(\tau)$
In this proof I do not see, what the point of the function $\iota$ is, and that you use
$\delta^{\mu_1}\wedge\dotso\wedge\delta^{\mu_k}(e_{\nu_1},\dotso, e_{\nu_k})=\iota^\ast\delta^{\mu_1}\wedge\dotso\wedge\iota^\ast\delta^{\mu_k}(e_{\nu_1},\dotso, e_{\nu_k})$
I can not see any step, where this is actually used, respectively could not done without it. For instance the last step.
Could I not simply write :
$\delta^{\mu_1}\wedge\dotso\wedge\delta^{\mu_k}(e_{\nu_1},\dotso, e_{\nu_k})=sgn(\tau)\delta^{\nu_1}\wedge\dotso\wedge\delta^{\nu_k}(e_{\nu_1},\dotso, e_{\nu_k})$
immediatly? Is there a difference?
Thanks in advance for any explanation to the use of $\iota:V_0\hookrightarrow V$.