Quantum plane is a bialgebra I am reading ‘Hopf algebras and their actions on rings’. Susan wrote the quantum plane as an example at 1.3.9 Example. He said $B = k \langle x,y \mid xy = qyx \rangle$, $0 \neq q \in k$ with coalgebra structure
$$
  \Delta(x) = x \otimes x \,,
  \quad
  \Delta(y) = y \otimes 1  + x \otimes y \,,
  \quad
  \epsilon(x) = 1 \,,
  \quad
  \epsilon(y) = 0 \,.
$$
But when I checked the bialgebra conditions, I met a question: What does $\Delta(xy)$ (or $\Delta(xx)$, $\Delta(yy)$) look like? I mean, the precise expression? Can the definition of $\Delta$ on $x$, $y$ induce $\Delta(xy)$?
 A: What Montgomery writes in their book needs to be understood as follows:


*

*There exists a unique homomorphism of $k$-algebras $Δ$ from $B$ to $B ⊗ B$ such that $Δ(x) = x ⊗ x$ and $Δ(y) = y ⊗ 1 + x ⊗ y$.

*There exists a unique homomorphism of $k$-algebras $ε$ from $B$ to $k$ such that $ε(x) = 1$ and $ε(y) = 0$.

*The $k$-algebra $B$, together with the two homomorphisms $Δ$ and $ε$, becomes a $k$-bialgebra (with $Δ$ serving as comultiplication and $ε$ as counit).


Therefore, $Δ$ and $ε$ are homomorphisms of $k$-algebras by their construction.
The same principle has already been used in previous examples:

*

*In Example 1.3.2, the action of $Δ$ and $ε$ is only specified on basis elements, i.e., on elements of the group $G$.
To apply both maps to arbitrary elements of the group algebra $kG$, the given formulas first need to be extended linearly.
(And it then needs to be checked that these linear maps are indeed homomorphisms of $k$-algebras.)

*In Example 1.3.3, the action of $Δ$ and $ε$ are only specified on elements of the Lie algebra $\mathfrak{g}$.
To apply both maps to arbitrary elements of the universal enveloping algebra $U(\mathfrak{g})$, the given formulas first need to be extended as homomorphisms of $k$-algebras.

