How is the antipode defined?

Question

In this page they define the antipode using the diagram

I am finding it difficult to understand what is going on. In partiulcar:

1. What is one the naked arrow going $k \to A$?

2. Shouldn't the top arrow go $S \otimes A : A \otimes A \to A \otimes A$? I mean, if we start with $a \in A$, we should go: $a \mapsto \Delta(a) = \sum_a a_{(1)} \otimes a_{(2)} \mapsto \sum_a S(a_{(1)}) \otimes a_{(2)}$ and that should equal $a \mapsto u(a)$ how?

Answer 1

1. Since $A$ is a $k$-algebra, the arrow $k \to A$ is the unit (sometimes written $\eta$) - the unique $k$-algebra morphism (it maps a scalar $t \in k$ to $t1 \in A$).

2. Since $A \otimes A$ is the coproduct (in $k$-algebras) of $A$ and $A$, the two maps $S, \operatorname{id} \colon A \to A$ define a unique map from the coproduct $A \otimes A \to A$. This is the map denoted $(S,\operatorname{id})$; it maps $a \otimes b$ to $S(a)b$.

So the statement that the square commutes is just that, if $a \in A$ and $\Delta(a) = \sum_i a_i \otimes b_i$, then $$ \sum_i S(a_i)b_i = \epsilon(a)1, $$ as explained on the following page.