# How to show that $\mathbb{K}\otimes_\mathbb{K}V\cong V$ using just the universal property of the tensor product.

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.

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 $$f$$.

• 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.
– Mark
Sep 23 '21 at 17:32
• @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. Sep 23 '21 at 17:53

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).

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.