Anti-homomorphism of Hopf Algebra I've got a quick question regarding the anti-homomorphism property. Specifically, what does it actually mean??
For a bit of context, I have the following question.
We define $U_q = U_q[o(3)]$ to be the associative algebra, with generators $L_0$, $L_+$ and $L_-$, with the defining relations; $[L_0, L_{\pm}] = \pm L_{\pm}$, $[L_+, L_-] = (q^{2L_0} - q^{-2L_0}) / (q - q^{-1})$.
The antipode has been defined such that $$S(I) = I$$ $$S(L_0) = -L_0$$  $$S(L_{\pm}) = -q^{-L_0} L_{\pm}q^{L_0} = -q^{\mp1}L_{\pm}$$
I'm being asked to show that $S$ constitutes an anti-homomorphism, for the $L_0$ and $L_{\pm}$ generators. However, I'm not really sure how to go about this, because I'm not too sure what the anti-homomorphism actually looks like, or really means.
Any insight into this would be great. Thank you!!
 A: If $A$ is an algebra, then anti-homomorphism $\phi:A\to A$ is a linear map such that $$\phi(x\cdot y)=\phi(y)\cdot\phi(x).$$ In your situation, you are probably being asked to show that there exists an anti-homomorphism $S$ which acts as you were told on the generators of the algebra.
A simple example, different from the one you want, so as not to ruin your problem:
Let $A$ be the algebra generated by two letters $x$ and $y$ subject to the relation $$yx-xy=1.$$ I claim there is a unique anti-homomorphism $\phi:A\to A$ such that $\phi(x)=y$ and $\phi(y)=-x$.
A: If $A$ is an associative algebra then its opposite algebra $A^{op}$ is defined on the same underlying vector space but with multiplication 
$$
x\cdot y=yx
$$
In the LHS, $x\cdot y$ stands for the multiplication in $A^{op}$, while on the RHS $yx$ stands for the original multiplication (in $A$). 
Then, an antihomomorphsism $\phi:A\to A$ is (by definition) an homomorphism  $\phi:A\to A^{op}$ from $A$ to the opposite algebra $A^{op}$. So, the antipode $S$ of a Hopf algebra $H$ is simply an algebra homomorphism $S:H\to H^{op}$. 
