I'm studying the construction of the tensor product. To do this, I opted for the quotient space's approach, defining first the free vector space over the cartesian $V\times W$ of two vector spaces. The following steps is at usual, take the quotient of the free vector space and the subspace spanned by especifics formal linear combinations etc. In this process I ended up by proving the universal propertie of free vector spaces:
For any set $X$ and vector space $V$, everymap $f\colon X\to V$ have an unique extension to a linear map $\overline{f}\colon\mathbb{K}^X\to V$, where $\mathbb{K}^X$ denotes the free $\mathbb{K}$-vector space over $X$.
Now, I'm a bit confused, because the universal propertie of the tensor product is:
The tensor product of two vector spaces $V$ and $W$ is a vector space denoted as $V\otimes W$, together with a map $\otimes\colon V\times W\to V\otimes W$ such that, every bilinear map $T\colon V\times W\to X$, there exists a unique linear map $\widetilde{T}$ such that $T = \widetilde{T}\circ \otimes$.
From what little I know about the universal property and category theory, both properties seems the same to me (equivalents) and this would implie that the free vector space of $V\times W$ and the tensor product $V\otimes W$ are related by a unique isomorphism. Now I don't think they are the same, since one can be constructed by the quocient of the another. So, where is my mistake?
One universal property is about extending a set-map to a linear map, and the other is about passing bilinear maps down to linear maps. These are very different universal properties, so there is no reason for $V \times W$ and $V \otimes W$ to be isomorphic.
It is true that two objects satisfying the same universal property are canonically isomorphic, but that is not your situation as you have two [different] properties in play, not one.