Exterior power of a space of maps $(\mathbb{K}^T)$ We are given a set $T \neq \emptyset, \ \ p \ge 1, \ \ p_i : T \rightarrow \mathbb{K}$
Could you help me prove that if
$ \phi: (\mathbb{K}^T)^p \ni (f_1, ..., f_p) \rightarrow \rho \in \mathbb{K}^{T^p}$ 
where $\rho: T^p \ni (x_1, ..., x_p) \rightarrow det [f_i(x_j)]_{i,j = 1, ... p} \in \mathbb{K}$
then $(\mathbb{K}^{T^p}, \phi)$ is the $p$-th exterior power of $\mathbb{K}^T$?
I know that $\phi$ is $p$-linear and anti-symmetric, because $\det$ is $p$-linear and anti-symmetric, but I have problems finding the unique linear map which makes the proper diagram commute.
Could you help me with that?
Thank you.
 A: Unfortunately, I couldn't easily see if the result is true for arbitrary $T$, so I will only answer for finite $T$. See the last part of the answer for the final word.
I use the notation in the comment of Michal Seweryn : $X = \langle \phi(V^p)\rangle$, with $V = K^T$. Then obviously since $\phi$ is $p$-linear alternating, we have the induced linear map $\tilde{\phi} : \Lambda^p V\to X$. We show that it is an isomorphism, which proves that $(X,\phi)$ is an exterior product by the usual yoga of universal properties. It is already clear that it is surjective, since $X$ is more or less by definition the image of $\tilde{\phi}$ in $K^{T^p}$.
Choose a set $S\subset T^p$ such that $S$ is a set of representatives for the action of the symmetric group $S_p$ of $\{(x_i)\in T^p | i\neq j\implies x_i\neq x_j\}$. If you want, you can say that $T = \{1,\dots,n\}$ and then take $S = \{(i_1,\dots,i_p)\in T^p | i_1<\dots <i_p\}$.
For any $x\in T$, we let $\delta_x\in V$ be the function $T\to K$ such that $\delta_x(y) = \delta_{x,y}$ (the usual delta function of Kronecker) ; and if $\overline{x}\in T^p$, then $\omega_{\overline{x}} = \delta_{x_1}\wedge\dots \wedge \delta_{x_p}\in \Lambda^p V$. Then $(\omega_{\overline{x}})_{\overline{x}\in S}$ is a basis of $\Lambda^p V$ (if $T = \{1,\dots,n\}$, then $(\delta_x)_{x\in T}$ is the usual canonical basis of $V$ and $(\omega_{\overline{x}})_{\overline{x}\in S}$ is the usual basis of $\Lambda^p V$).
I view $K^{T^p}$ as the space of function $T^p \to K$. I then define $\psi: X\to \Lambda^p V$ by $\psi(f) = \sum_{\overline{x}\in S} f(\overline{x})\omega_{\overline{x}}$.
Let's check that it is the inverse of $\tilde{\phi}$. Note that by definition, $\tilde{\phi}(f_1\wedge\dots \wedge f_p) = \left( (x_j)\mapsto \det(f_i(x_j)) \right)$.
We have $(\psi\circ \tilde{\phi}) (\omega_{\overline{x}}) = \sum_{\overline{y}\in S} \det(\delta_{x_i}(y_j)) \omega_{\overline{y}}$. Now it is easy to see that by our choice of $S$ $\det(\delta_{x_i}(y_j))$ is $1$ if $\overline{x}=\overline{y}$ and $0$ otherwise. So $(\psi\circ \tilde{\phi}) (\omega_{\overline{x}}) = \omega_{\overline{x}}$.
Conversely, if $f\in X$, then $(\tilde{\phi}\circ \psi)(f)$ sends $\overline{x}\in S$ to 
$\begin{eqnarray*}
 & & \tilde{\phi}\left( \sum_{\overline{y}\in S} f(\overline{y})\omega_{\overline{y}} \right) (\overline{x}) \\
 & = & \sum_{\overline{y}\in S} f(\overline{y})\cdot \tilde{\phi}(\omega_{\overline{y}})(\overline{x}) \\
 & = & \sum_{\overline{y}\in S} f(\overline{y}) \det(\delta_{y_i}(x_j)) \\
 & = & f(\overline{x})
\end{eqnarray*}$
by the same remark as above converning $\det(\delta_{y_i}(x_j))$. But now, it is easy to see that any funtion in $X$ is uniquely determined by its values on the $\overline{x}$, because by construction if $f\in X$ then $f(\sigma\cdot \overline{x}) = \varepsilon(\sigma) f(\overline{x})$ where $\sigma\in S_p$ acts naturally on $T^p$ and $\varepsilon(\sigma)$ is the signature. So $(\tilde{\phi}\circ \psi)(f) = f$.
In the end, $\tilde{\phi}$ and $\psi$ are indeed inverse linear maps. 
We even see that they exchange the bases $(\omega_\overline{x})_{\overline{x}\in S}$ and $(f_\overline{x})_{\overline{x}\in S}$, where $f_{\overline{x}} = \sum_{\sigma\in S_p} \varepsilon(\sigma)\delta_{\sigma\cdot \overline{x}}$ and $\delta_{\overline{x}}$ is the usual Kronecker function $T^p\to K$ (we could alternatively have defined the $f_\overline{x}$, showed that they were a basis of $X$ and that $\tilde{\phi}$ has the expected action).

So if $m: V^p \to W$ is any $p$-linear alternating map, what is the induced linear map $\tilde{m}: X\to W$ ?
Using the above isomorphisms, we can see that if $f\in X$, $\tilde{m}(f) = \sum_{\overline{x}\in S}f(\overline{x})m(\overline{x})$. 
Now if he characteristic of $K$ is strictly greater than $p$, there is a nicer formula that does not depend on the choice of $S$. Indeed, 
$\sum_{\overline{x}\in T^p}f(\overline{x})m(\overline{x}) = \sum_{\overline{x}\in S}\sum_{\sigma\in S_p}f(\overline{\sigma\cdot x})m(\overline{\sigma\cdot x}) = p! \sum_{\overline{x}\in S}f(\overline{x})m(\overline{x})$, so when $p!$ is invertible we get 
$$\tilde{m}(f) = \frac{1}{p!}\sum_{\overline{x}\in T^p}f(\overline{x})m(\overline{x}).$$
