On proposition I.3.2 of 'Quantum groups' by Kassel.

Question

I am reading the book Quantum groups by Kassel. In proposition I.3.2 at the very beginning the reader is asked to show that under the identifications made, the maps $\Delta,\varepsilon$ and $S$ correspond to the maps $+,0$ and $-$.

However, $+$ is a map from $A^2$ to $A$, and I'm not sure how we can use the identifications to see that $\Delta:k[x]\rightarrow k[x',x'']$ corresponds to $+$.

It seems to me, all you can get from the identifications is that the map $\Delta$ corresponds to the element $x'+x''\in k[x',x'']$. Similarly, $\varepsilon$ corresponds to the element $0\in k$ and $S$ corresponds to the element $-x\in k[x]$.

So what exactly is meant in this proposition? I'm fairly sure this is a stupid question, but it's one that should be well-understood before proceeding any further in this theory.

Thank you in advance.

Answer 1

$\text{Hom}(k[x], -)$ is the forgetful functor from $k$-algebras to sets; that is, a homomorphism $k[x] \to k[x', x'']$ of $k$-algebras is the same thing as an element of $k[x', x'']$, namely the image of $x$. Alternatively, the point is that a morphism $A \times A \to A$ of affine varieties over $k$ (here $A$ is the affine line) dualizes to a morphism $k[A] \to k[A] \otimes k[A]$ in the opposite category, and $k[A] \cong k[x]$ while $k[A] \otimes k[A] \cong k[x', x'']$. The other cases are similar.

This can be used to set up various equivalences; for example, you get a contravariant equivalence of categories between commutative Hopf algebras over $k$ and affine group schemes over $k$.

Answer 2

The functor $\mathrm{Hom}(-,A)$ is a contravariant functor, so a map $f:k[x]\to k[x',x'']$ corresponds to a map $\mathrm{Hom}(k[x',x''],A)\to\mathrm{Hom}(k[x],A)$ given by pre-composing with $f$. If you write down what this means for $\Delta$ you start with a map $k[x',x'']\to A$ that takes for instance $x'\mapsto a$ and $x''\mapsto b$, corresponding to $(a,b)\in A^2$. Pre-composing with $\Delta$ we get the map $x\mapsto a+b$, which corresponds to $a+b\in A$. So we see that $\Delta$ corresponds to the map $(a,b)\mapsto a+b$. You can do the same to get the correspondences for $S$ and $\epsilon$.

Answer 3

Recently, I read the book by Kassel. What disturbed me about his presentation that it was not clear how to come up with the maps $\Delta, \epsilon$ and $S$ starting from the addition maps $+ : A^2 \to A$ from your algebras. However, once you think about this, the Yoneda lemma comes to the rescue. Indeed, given a commutative algebra $A$, we can consider the map

$$\text{Hom}_{\text{Alg}}(k[x',x''],A) \cong A^2 \stackrel{+}\longrightarrow A \cong \text{Hom}_{\text{Alg}}(k[x],A)$$

This gives a natural transformation between the functors $$\text{Hom}_{\text{Alg}}(k[x',x''],-) \to \text{Hom}_{\text{Alg}}(k[x],-)$$ which by the Yoneda lemma gives an algebra morphism $$\Delta: k[x]\to k[x',x'']\cong k[x]\otimes k[x]$$

One then calculates that $\Delta$ is the desired algebra morphism, i.e. that $\Delta(x) = x' + x''$. Under the canonical identification $k[x',x'']\cong k[x]\otimes k[x]$, we then obtain $$\Delta(x) = x \otimes 1 + 1 \otimes x.$$