Proposition 5

(i) Let E be a graded cogebra admitting a counit $\gamma$; then $\gamma$ is a homogeneous linear form of degree 0.

(ii) Suppose further that there exists an element $e \in E$ such that $E_0 = Ae$ and $\gamma(e) = 1$. Then the kernel $E_\gamma$ of $\gamma$ is equal to $E_+ = \sum_{n \geq 1} E_n$, $c(e) = e \otimes e$ and

for all $x \in E_+$.

(i) It suffices to verify that $\gamma(x) = 0$ for $x \in E_n$, for all $n \geq 1$. Since $c$ is a graded homomorphism of degree 0,

with, for all $j$ such that $0 \leq j \leq n$, $y_{ij}$ and $z_{ij}$ in $E_j$; applying (16) (no. 2) we obtain

whence, equating the components of degree 0 and degree $n$ on the two sides

and therefore $\gamma(x) = \sum_i \gamma(y_{in}) \gamma(z_{i0}) = \gamma(0) = 0$.

(ii) Since $\mathrm{Ker}(\gamma)$ and $E_+$ are both supplementary sub-A-modules of $Ae = E_0$ and $E_+ \subset \mathrm{Ker}(\gamma)$ by (i), $E_+ = \mathrm{Ker}(\gamma)$ (II, § 1, no. 8, Remark 1); the other assertions follow from Proposition 4 of no. 2.