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If I’m given a finite-dimensional Hopf algebra $H$, how do I show the antipode is bijective? It's obvious that if we prove either injective or surjective, we get the other one for free since $H$ is finite-dimensional. Can someone nudge me in the right direction?

Jendrik Stelzner
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fosterc4
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    Googling makes it seem like some places require $S$ to be invertible, and others don't which in theory could lead to examples where it isn't. – anon Mar 18 '21 at 22:50
  • I did manage to find a proof here: https://www.famaf.unc.edu.ar/~andrus/papers/Schn1.pdf It requires a lot of tools though. – fosterc4 Mar 19 '21 at 19:54

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This is a basic fact in the theory of Hopf algebras. It is a corollary of the fundamental theorem of Hopf modules. Reference : Montgomery Susan, Hopf algebras and their actions on rings, Theorem 2.1.3.

Functor
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    While the reference you give may provide an answer to the Question, it is the recommended approach to quote or summarize the most germane material to be found there. This gives your Readers a better idea whether it would be worth pursuing the reference. A Commenter (the OP) indicated the result "requires a lot of tools," so perhaps a good Answer would sketch which topic and preliminary ideas are needed for this. – hardmath Jan 30 '22 at 15:45