Given vector spaces over a field $\mathbb{K}$, how can one show that $\mathbb{K}\otimes_\mathbb{K}V\cong V$ using just the universal property of the tensor product? Some background: This is an exercise in Riehl's book "Categories in Context" that I have gotten stuck on. I am not very familiar with tensor products so I am not sure if my struggling has to do with that unfamiliarity or not understanding universal properties.
I have tried constructing a map from $\mathbb{K}\otimes_\mathbb{K}V$ to $V$ and then showing it has an inverse. The map I tried is the map $\bar{P}:\mathbb{K}\otimes_\mathbb{K}V\rightarrow V$ induced by the projection map $P:\mathbb{K}\times V\rightarrow V$. The map $\bar{P}$ comes from the isomorphism $Vect(\mathbb{K}\otimes_\mathbb{K}V,-)\cong Bilin(\mathbb{K}, V;-)$. I can show that $\bar{P}\circ(\otimes\circ i)=1_V$, where $i$ is a section of $P$. But I cannot show there is a left inverse to $\bar{P}$. Any suggestions or tips would be greatly appreciated.
Best Answer
The tensor product can be defined by its universal property, so to prove that something ($V$ here) is a tensor product using just the universal property, one has to exhibit the universal property for that something ($V$) and invoke the statement that the object with the said universal property is unique.
The defining property of the tensor product $A \otimes B$ is that any map bilinear map $A \times B \to X$ factors through it via the canonical map $A \times B \to A \otimes B.$
So you have to assume that you have a bilinear map $f: K \times V \to X$ and prove that it factors through some canonical map as $K \times V \to V \to X.$ Where can one send a $(k,v)$ to get an element of $V$? The most natural choice is $kv.$ And indeed the bilinearity of $f$ implies that $f(k,v) = f(1,kv),$ so you can first pair scalars with vectors, and only then perform mapping by $f$.