Bialgebra

Remarks

Clause 3 means $\Delta(ab) = \Delta(a)\Delta(b)$, $\Delta(1) = 1 \otimes 1$, $\varepsilon(ab) = \varepsilon(a)\varepsilon(b)$ and $\varepsilon(1) = 1$. It is equivalent to asking that $m$ and $\eta$ be coalgebra homomorphisms. Example: the group algebra $KG$ of a group $G$, with $\Delta(g) = g \otimes g$ and $\varepsilon(g) = 1$ for $g \in G$. The check of clause 3 on the basis: $\Delta(gh) = gh \otimes gh = (g \otimes g)(h \otimes h) = \Delta(g)\Delta(h)$ and $\varepsilon(gh) = 1 = \varepsilon(g)\varepsilon(h)$. Non-example: on $K[x]$, the coalgebra with $\Delta(x^n) = \sum_{k=0}^{n} x^k \otimes x^{n-k}$ and $\varepsilon(x^n) = \delta_{n,0}$ fails clause 3. Indeed $\Delta(x)^2 = (x \otimes 1 + 1 \otimes x)^2 = x^2 \otimes 1 + 2\, x \otimes x + 1 \otimes x^2$, while $\Delta(x^2) = x^2 \otimes 1 + x \otimes x + 1 \otimes x^2$, and $2 \ne 1$ in every field. With $\Delta(x^n) = \sum_{k} \binom{n}{k} x^k \otimes x^{n-k}$ instead, $K[x]$ is a bialgebra.