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Let $A$ be a hopf algebra associated to some linear algebraic group.

I read that: \begin{array}{ccc}A&\overset{id\otimes\iota}\longleftarrow&A\otimes A\\\iota\otimes id\uparrow&&\uparrow \Delta\\A\otimes A&\overset{\Delta}{\longleftarrow}&A\end{array}

must be satisfied, where $\iota$ is the coinverse. I am unsure why these live in the places that they do. Shouldn't $\iota:A\to A$ and $id:A\to A$ mean that $\iota\otimes id:A\otimes A \to A\otimes A$, why is the last square just $A$?

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    The tensor in the mappings here should be read as a multiplication. So, $a \otimes b$ is sent by $\iota \otimes \mathrm{id}$ to $\iota(a)b$. – 2017-02-21
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    @MeesdeVries That would fix my concern. How do I distinguish when that is the case? I.e. $\Delta \otimes id$ should definitely map $A\otimes A\to A\otimes A\otimes A$ right? – 2017-02-21
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    I can see why $\epsilon \otimes id$ would map $A\otimes A\to A$, since it'll take the left term into the field, but in this case I just don't see it. – 2017-02-21
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    I think the notation in the diagram is just bad. It should explicitly mention a multiplication mapping somewhere. But in this case you can tell what they mean by looking at the domain/codomain. – 2017-02-21
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    @MeesdeVries Is this coinverse actually the antipode? I am trying to compare to other notes, and it seems like it might be – 2017-02-21
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    Yes, those are synonyms. (At least as far as I know.) – 2017-02-21
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    @MeesdeVries Thanks very much! – 2017-02-21

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