Hopf Algebra Structure of $U_q(\mathfrak {sl}(2))$. I am working out of Christian Kassel Quantum Groups.
Define $U_q=U_q(\mathfrak {sl}(2))$ as the algebra generated by elements $E,F,K,K^{-1}$ subject to the following relations.
$$
  K K^{-1} = K^{-1} K = 1 \,,
  \quad
  K E K^{-1} = q^2E \,,
  \quad
  K F K^{-1} = q^{-2} F \,,
  \quad
  [E, F] = \frac{K - K^{-1}}{q - q^{-1}} \,.
$$
Now define the following maps $\Delta$, $\epsilon$, $S$:
\begin{gather*}
  \Delta(E) = 1 \otimes E + E \otimes K \,,
  \quad
  \Delta(F) = K^{-1} \otimes F + F \otimes 1 \,,
  \quad
  \Delta(K^{\pm 1}) = K^{\mp 1} \otimes K^{\mp 1} \,,
  \\[0.5em]
  \epsilon(E) = \epsilon(F) = 0 \,,
  \quad
  \epsilon(K) = \epsilon(K^{-1}) = 1 \,,
  \\[0.5em]
  S(E) = -EK^{-1} \,,
  \quad
  S(F) = -KF \,,
  \quad
  S(K^{\pm 1}) = K^{\mp 1} \,.
\end{gather*}
I am trying to verify that this is a Hopf algebra structure on $U_q$. Everything works out fine until I try to verify that $S$ defines an antipode. By Kassel’s own definition earlier in the text an antipode on a bialgebra is an endomorphism such that the following holds:
$$
  S \star \iota = \iota \star S = \eta \circ \epsilon \,.
$$
Initially I began by trying to verify that $S$ was an endomorphism. If so, the following should hold:
$$
  S(K) S(E) S(K^{-1}) = q^2 S(E) \,.
$$
The problem is that when you sub in the given values I get that
$$
  K^{-1} E = q^2 E K^{-1} \,,
$$
and as far as I can tell there is no way to rewrite this as one of the given relations.
As a side note for some reason Kassel says that he wants $S$ to be a morphism from $U_q$ to $U_q^{\mathrm{op}}$ but I didn’t understand quite what this was or how this fit in with his own defintion.
 A: As said in the comments, $S$ is an anti-algebra morphism meaning that $S(xy) = S(y)S(x)$ for all $x,y \in U$. For example, to see that $S$ is compatible with $KEK^{-1} = q^2E$ one needs to check that $S(K^{-1})S(E)S(K) = q^2S(E)$ i.e $K EK^{-1}K^{-1} = q^2EK^{-1}$ which is exactly the relation $KEK^{-1} = q^2E$. 
For the second part let us check e.g that $S * i = \eta \circ \varepsilon$. It's enough to look at generators $K,E$ and $F$. Recall that for endomorphisms $f,g$ of a Hopf algebra we have by definition $f*g(x) := \mu \circ (f \otimes g) \circ \Delta(x)$. 
To conclude we just need to check $S*i(x)= \eta \circ \varepsilon(x)$ in the three cases $x \in \{K,E,F\}$ : $$(S*i)(K) = \mu \circ ((S \otimes i)(K \otimes K)) = \mu(K^{-1} \otimes K) = 1$$ $$\eta \circ \varepsilon(K) = \eta(1) = 1$$
$$(S*i)(E) = \mu (S(1) \otimes E + S(E) \otimes K) = \mu (1 \otimes E - EK^{-1} \otimes K) = E - E = 0$$ $$\eta \circ \varepsilon (E) = \eta(0) = 0$$
$$(S*i)(F) = \mu (S(K^{-1}) \otimes F + S(F) \otimes 1) = \mu (K \otimes F - KF \otimes 1) = KF - KF = 0$$ $$\eta \circ \varepsilon (F) = \eta(0) = 0$$
