In what follows, $U$ is a complex vector space. I have the following abstract definition.
Definition: Let $\wedge^{n}: \overbrace{U \times \cdots \times U}^{\text{$n$ times}} \to V$ be an alternating $p$-linear map. We say that $\wedge^{p}$ satisfies the universal property if given another alternating $p$-linear map $f: U \times \cdots \times U \to W$, then there exists a unique map $h: V \to W$ such that $f = h\circ \wedge^{n}$.
Now, suppose $U$ is finite dimensional. Consider $V = \text{Alt}^{n}(U)$ the space of alternating tensors and let:$$\wedge^{n}(v_{1},...,v_{n}):= \frac{1}{n!}\sum_{\sigma \in S_{n}}\text{sign}(\sigma)v_{\sigma(1)}\otimes \cdots \otimes v_{\sigma(n)}$$
This map is clearly alternating an $p$-linear. My question is: does it also satisfy the universal property? I couldn't prove it myself.
