Finding the inverse of a map from $\wedge^{k}\Bbb V^*\to \text{Hom}(\wedge^{n-k}\Bbb V,\wedge^n\Bbb V^*)$. I am new to differential geometry and I have encountered a problem regarding $k$-forms and multilinear algebra.
Let $\Bbb V$ be a vector space of dimension $n$ and let $0\leq k\leq n$.
For any $\alpha\in\wedge^{k}\Bbb V^*$, consider the map from $\wedge^{k}\Bbb V^*\to \text{Hom}(\wedge^{n-k}\Bbb V^*,\wedge^n\Bbb V^*)$ sending $\alpha\mapsto A_\alpha$
where $A_\alpha(\beta)=\alpha\wedge\beta$ for $\beta\in\wedge^{n-k}\Bbb V^*$.
I have shown that this map is an isomorphism, but how to compute the inverse in a non-messy way?
 A: It's not possible without choosing a basis - in this case you are asking about the inverse to the map:
$$\bigwedge ^k V \to \bigwedge ^{n-k} V^*$$
If we could do this for any $k$ and any $V$, then by applying the same technique to $\bigwedge ^{n-k} V^*$ on the left hand side, we could get an inverse for the composition:
$$\bigwedge ^k V \to \bigwedge ^{n-k} V^* \to \bigwedge ^k V^{**}$$
In the case of $k=1$ this is just the natural map $V\to V^{**}$. Since I don't believe there is any "natural" way to write down the inverse of the isomorphism with the double dual (see background below), it seems there can be no way to do it for this more general problem either.

Background. There isn't always a "non-messy" way to find an inverse for natural maps that involve duals of vector spaces - one often must take a basis to construct it, although of course the answer does not depend on the choice.


*

*For instance, take the map $$V \to V^{**}$$ which is completely natural and an isomorphism for finite dimensional $V$. To define an inverse, one takes a basis $\{e_i\}$ for $V$ then send $\omega \mapsto \sum \omega(f_i)e_i$ where $f_i(e_j) = \delta_{ij}$ is a dual basis. Perhaps this is unnatural because it doesn't work for all vector spaces - only finite dimensional ones. Otherwise the map is not an isomorphism, and if you try to do the same thing as above, the sum will in general be an infinite sum, which is undefined in the language of vector spaces.

*Another example is the isomorphism $\phi:$ $$V^* \otimes V^* \to (V\otimes V)^*$$ defined on pure tensors by $\phi(f,g)(v\otimes w) = f(v)\cdot g(w)$. It is a natural map, but the inverse again is not so natural - one usually picks a basis $\{e_i\}$, builds the dual basis $\{f_i\}$ satisfying $f_i(e_j) = \delta_{ij}$, and then checks that $f_i\otimes f_j$ is a dual basis for $\{e_i\otimes e_j\}$ to see $\phi$ is an isomorphism.
So often when duals are involved, many maps are naturally defined for all vector spaces, but fail to be isomorphisms for infinite dimensional problems. In these cases, it should not be possible to write a formula for the inverse without picking a basis, even though any basis is an acceptable choice, and the formula you get does not depend on the choices made.

How to do it with a basis. If you don't mind using a basis, you should pick an ordered basis $e_1,\ldots, e_n$ for $V$ with dual basis $f_i$ such that $f_i(e_j) = \delta_{ij}$. Then the basis of $\bigwedge ^k V$ is $e_{i_1}\wedge\cdots\wedge e_{i_k}$ for $i_1<\cdots < i_k$ and a basis of $\bigwedge ^{n-k}V^*$ is the wedges $f_{j_1}\wedge \cdots \wedge f_{j_{n-k}}$ in a similar way.
Given the (ordered) subset $\{j_1,\ldots, j_{n-k}\} \subset \{1,\ldots, n\}$ there is a complimentary subset $\{i_1,\ldots, i_k\} = \{1,\ldots, n\} \setminus \{j_1,\ldots, j_{n-k}\}$ so you can send a pure tensor
$$f_{j_1}\wedge \cdots \wedge f_{j_{n-k}} \mapsto e_{i_1}\wedge \cdots \wedge e_{i_k}$$
and this will equal the inverse map.
