Proving uniqueness of antipodes in Hopf algebras

Question

Let $(H,\mu,\nu,\Delta,\epsilon)$ be a Bialgebra where H is the vector space, $\mu, \nu$ are the product and unit whilst $\Delta, \epsilon$ are the coproduct and counit. Now, for $f,g \in end(H)$ define $f@g \in end(H)$ by $f@g=\mu(f \otimes g)\Delta(x)=\Sigma_{(x)}f(x')g(x'')$ (via Sweedler notation).

An element $S \in end(H)$ is called an antipode if

$S@id_H=id_H@S=\nu\circ\epsilon$

If a Bialgebra has an antipode, then it is unique. To see this, suppose $S,T$ are antipodes for the bialgebra $H$. Then we have:

$S = S@(\nu\epsilon)=S@(id_H@T)=(S@id_H)@T=(\nu\epsilon)@T=T$

Can somebody explain to me the first equality? Why do we get $S = S@(\nu\epsilon)$?

Answer 1

We know that an antipode for a bialgebra $H$ is a linear map $S:H \rightarrow H$ which is a convolution inverse of $id_H$.

We have to remember that the unit for the convolution product is $$1_{\text{End}(H)}=\nu \circ\epsilon $$ Proof: $$f*(\nu\epsilon) = \mu \circ(f \otimes \nu\epsilon) \circ \Delta = \mu \circ (A \otimes \nu) \circ (f \otimes \mathbb{k})\circ (C \otimes \epsilon) \circ \Delta = \\= r_A \circ (f \otimes \mathbb{k}) \circ(r_C)^{-1} = f$$ (if you miss the second equality you can read a complete explanation here)

So, $\forall f \in \text{End}(H)$, the following holds (because $f*id = f$ in every algebraic context): $$f * 1_{\text{End}(H)}= f$$

Now, take $S \in \text{End}(H)$ and write the same thing: $$S = S * 1_{\text{End}(H)}$$

but $1_{\text{End}(H)}=\nu \circ\epsilon $, so

$$S = S *\nu \epsilon$$

Answer 2

We can in fact prove this using a commutative diagram, so that the result generalizes for all Hopf monoids in symmetric monoidal categories.

Here, $I$ denotes the monoidal unit (that is, the base field). The two top quadrilaterals commute by the properties of a monoidal category; the two triangles commute by the properties of the unit $\eta$ and counit $\varepsilon$; the bottom quadrilateral commutes because $T$ is an antipode. Therefore, we have proven that:

$$S=(\mu\circ(1_H\otimes\mu))\circ(S\otimes1_H\otimes T)\circ((1_H\otimes\Delta)\circ\Delta).$$

In a very similar manner we can also prove that

$$T=(\mu\circ(\mu\otimes1_H))\circ(S\otimes1_H\otimes T)\circ((\Delta\otimes1_H)\circ\Delta).$$

But by associativity and co-associativity, this means $S=T$.