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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.

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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 $f$.

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  • $\begingroup$ Proving using the universal property means that you should use it to define the isomorphism, not necessary to show that $V$ itself has this universal property. Your answer is still correct though. $\endgroup$
    – Mark
    Sep 23 '21 at 17:32
  • $\begingroup$ @Mark Interesting: that's the opposite of what I would call a genuine proof by universal property. Such a proof should not use the explicit form of the universal thing. $\endgroup$ Sep 23 '21 at 17:53
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Hint: It's enough to prove that $V$ itself satisfies the universal property that any bilinear map $K\times V\to W$ factors through it.
More specifically, they should uniquely factor through the bilinear map $K\times V\to V$ which maps $(k,v)\mapsto k\cdot v\,$ (the projection is not bilinear).

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Define the map $f:V\to K\otimes_K V$ by $v\to 1\otimes v$. Also, let $g: K\times V\to V$ be the map $g(k,v)=kv$. It is easy to show that $g$ is bilinear, and so by the universal property it induces a linear map $\tilde{g}:K\otimes_K V\to V$. Now just check that $f$ and $\tilde{g}$ are inverses of each other, I'll leave it to you.

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