Let $V$ be a finite-dimensional vector space over $\Bbb Q$ and $\bigwedge^kV$ be the $k$-th antisymmetric power. Let $\{ v_1, \dots, v_n \}$ be a basis of V. Define $$\pi: V^{\otimes k} \to \bigwedge^kV, \\ v_{i_1}\otimes \dots \otimes v_{i_k} \mapsto v_{i_1}\land \dots \land v_{i_k}$$ and $$\iota: \bigwedge^kV \to V^{\otimes k}, \\ v_{i_1} \land \dots \land v_{i_k} \mapsto \sum_{\sigma \in S_k}\mbox{sgn}(\sigma)~v_{\sigma(i_1)}\otimes \dots \otimes v_{\sigma(i_k)}$$ Where $S_k$ is a symmetric group in $K$ elements.
a) Show that $\pi$ and $\iota$ are both $\Bbb K$- linear maps.
b) Show that $\pi$ is surjective and $\iota$ is injective.
c) Determine $\iota \circ \pi$.
For a) I was thinking maybe I could define a map $$\phi: V^k \rightarrow \bigwedge^k(V) \\ (v_1,...,v_n) \mapsto v_1\land v_2 \land \cdot \cdot \cdot \land v_n$$ which is clearly multilinear and then by the universal property there is a linear map: $$\bar{\phi}: v_1 \otimes \cdots \otimes v_k \rightarrow V^{\otimes k}$$ Since the composition of a multilinear map and a linear map is multilinear : $\pi =\phi \circ \bar{\phi}$.
Fir $\iota$ i have a similar method in mind.
Can someone tell me if this is correct and if not, maybe some hints?
c) I want to give the answer that it takes $\bigwedge^k V $ to $\bigwedge^k V $, but that seems to be such an easy answer i feel like thats not what the exercise wants me to show,
and as for b) can someone give me a few hints?
For b) I have an idea how to show $\iota$ is injective, but that is a property of the permutation, is that applicable here? So we show that the map is symmetric and then by the property of permutation its injective (I'm typing on my Ipad, typing everything out in Latex would take too long)..
Definition $V^r{\otimes}$: for every natural number $r$ the r-th tensor power of $V$ is defined to be $V^{\otimes 0}=K$ - $V^{\otimes r}:V \otimes_{K}…\otimes_K V$ with $r>0$
For $\land V $ it’s the usual definition from exterior algebra