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Let $(H, \nu,\eta, \Delta, \epsilon, S)$ be a Hopf algebra. $S$ is the antipode.

I am reading a proof of the fact $S(xy)=S(y)S(x)$.

First, define maps $\nu, \rho$ in $\hom(H \otimes H, H)$ by

$\nu(x \otimes y)=S(y)S(x)$ and $\rho(x \otimes y)=S(xy)$.

To prove the fact, we have to show that $\rho=\nu$. It is written that to show this it suffices to show that $\rho* \mu=\mu*\nu=\eta\epsilon$, where $*$ is the convolution.

I don't understand why this implies that $\rho=\nu$.

Please give some advice.

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In an associative algebra, if an element has a left inverse and a right inverse, these two are equal.

Now $\rho\star\mu=\eta\epsilon$ and $\mu\star\nu=\eta\epsilon$ mean precisely that $\rho$ and $\nu$ are left and right inverse to $\mu$, respectively, in the convolution algebra $\hom(H\otimes H,H)$ (recall that $\eta\epsilon$ is the unit element in that algebra)

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Firstly, the unit of the convolution algebra $Hom(H\otimes H,H)$ is $\iota(x\otimes y) = \epsilon(x)\epsilon(y)1_H$.

Now, we can prove that $m$ is inverse of $\rho$, i.e $\rho \ast m = \iota$.

$$\rho\ast m (x\otimes y) = \sum_x \sum_y S(x^{(1)}y^{(1)})x^{(2)}y^{(2)} = \sum_{xy} S((xy)^{(1)})(xy)^{(2)} = \epsilon(xy)1_H = \iota(x\otimes y)$$

Next step, we prove that $m\ast \nu = \iota$.

$$m\ast \nu (x\otimes y) = \sum_x\sum_y x^{(1)}y^{(1)}S(y^{(2)})S(x^{(2)}) = \epsilon(y)\sum_{x}x^{(1)}S(x^{(2)})=\epsilon(y)\epsilon(x)1_H = \iota(x\otimes y).$$

So, we get $\rho = \rho \ast (m\ast \nu) = (\rho \ast m)\ast \nu = \nu$.

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