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In some books there are assertions, usually without proof, that $m \circ (S \otimes S) = S \circ m \circ\tau : A^{\otimes 2} \rightarrow A$ and $\tau \circ (S \otimes S)\circ \Delta = \Delta \circ S : A \rightarrow A^{\otimes 2}$ Where $\Delta$ coproduct, $\tau$ is switching morphism, $S$ is antipode, $m$ is multiplication. It is said that it follows from Hopf algebra axioms. How..?

A.Babichenko
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  • the first is the statement that S is an antihomomorphism, which you can find in any Hopf algebra source e.g. https://books.google.co.uk/books?id=pBJ6sbPHA0IC&pg=PA153&dq=%22is+an+antimorphism+of+algebras%22&redir_esc=y#v=onepage&q=%22is%20an%20antimorphism%20of%20algebras%22&f=false – Matthew Towers May 12 '21 at 12:03
  • the second is a dupe of https://math.stackexchange.com/questions/1878942/commultiplication-and-antipode-in-hopf-algebra – Matthew Towers May 12 '21 at 12:11
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    Proofs of both statements can be found in Kassel's Quantum Groups as Theorem III.3.4(a). – Elliot Yu May 18 '21 at 16:33

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