before I ask my question, I briefly review two definition which I use:
(1) Tensor product of vector spaces:
Let $V_{1},\dots,V_{n}$ be vector spaces over the same field $\mathbb{F}$. Then the ''tensor product'' $V_{1}\otimes\dots\otimes V_{n}$ is defined as the vector space with the following property: There exists a n-linear function $\otimes:V_{1}\times\dots\times V_{n}\to V_{1}\otimes\dots\otimes V_{n}$ such that for all vector spaces $W$ over $\mathbb{F}$ and for all n-linear functions $\tau:V_{1}\times\dots\times V_{n}\to W$, there exists a unique linear function $\psi:V_{1}\otimes\dots\otimes V_{n}\to W$ such that $\tau=\psi\circ\otimes$.
(2) exterior power
Let $V$ be a vector field over $\mathbb{F}$. Then the n-th exterior power is defined as $${\bigwedge}^{n}V:=\underbrace{V\otimes\dots\otimes V}_{n-times}/\operatorname{span}\{v_{1}\otimes\dots\otimes v_{n}\mid v_{1},\dots,v_{n}\in V\land\exists i,j\in\{1,\dots,n\}: v_{i}=v_{j}\}$$ Furthermore, we define the wedge product $\wedge:\underbrace{V\times\dots\times V}_{n-times}\to {\bigwedge}^{n}V$ through $\wedge:=\pi\circ\otimes$, where $\pi:\underbrace{V\otimes\dots\otimes V}_{n-times}\to {\bigwedge}^{n}V$ is the quotient map.
My goal is to prove that also the exterior power fulfils a similar universal property than the tensor product:
Theorem: For all vector spaces $W$ over $\mathbb{F}$ and for all alternating $n$-linear functions $\tau:\underbrace{V\times\dots\times V}_{n-times}\to W$ there exists an unique linear map $\psi:{\bigwedge}^{n}V\to W$, such that $\tau=\psi\circ\wedge$.
My first idea was to use the definition of the tensor product. Namely, let $W$ be a vector space over $\mathbb{F}$ and $\tau:\underbrace{V\times\dots\times V}_{n-times}\to W$ an alternating $n$-linear map. Then we know that there exists an unique linear map $\varphi:\underbrace{V\otimes\dots\otimes V}_{n-times}\to W$ such that $\tau=\varphi\circ\otimes$. Then I want to define $\psi:=\varphi\circ\pi^{-1}$, because then I would have $\tau=\psi\circ\wedge$. However, the problem is that the quotient map $\pi$ is not bijective and therefore $\pi$ has no inverse.....Any other proof ideas?