Proposition 4

Let E be a cogebra admitting a counit $\gamma$ and suppose that there exists in E an element e such that $\gamma(e) = 1$; then E is the direct sum of the sub-A-modules Ae and $E_{\gamma} = \mathrm{Ker}(\gamma)$ and

The first assertion is immediate, for $\gamma(x - \gamma(x)e) = 0$ and the relation $\gamma(\alpha e) = 0$ implies $\alpha = 0$. Let $c(e) = \sum_i s_i \otimes t_i$, so that

by (16) and $1 = \gamma(e) = \sum_i \gamma(s_i)\gamma(t_i)$. Therefore which, by the above relation, is just $c(e) - e \otimes e$; this therefore proves the first relation of (18). On the other hand the decomposition of $E \otimes E$ as a direct sum allows us to write, for $x \in E_\gamma$, $c(x) = \lambda(e \otimes e) + (e \otimes y) + z \otimes e) + u$ where $u = \sum_j v_j \otimes w_j, y, z$ and the $v_j$ and $w_j$ belong to $E_\gamma$. The definition of the counit $\gamma$ then gives $x = \lambda e + y = \lambda e + z$ and, as $\gamma(x) = 0$, necessarily $\lambda = 0, x = y = z$, whence the second relation of (18).