Proposition 3

Let E be a cogebra over A. In order that, for every unital A-algebra B, the A-algebra $\mathrm{Hom}_A(E, B)$ be unital, it is necessary and sufficient that there exist a linear form $\gamma$ on E rendering commutative the diagrams

$h'$ $h''$

where $c: E \to E \otimes_A E$ is the coproduct and $h'$ and $h''$ the canonical isomorphisms

(II, § 3, no. 4, Proposition 4). The unit of Hom$_A$(E, B) is then the linear mapping x $\mapsto$ $\gamma(x)1$ (where 1 denotes the unit element of B).

Let $\gamma$ be a linear form on E rendering diagram (16) commutative. Let B be a unital A-algebra with unit element 1, $\eta:A \to B$ the canonical mapping and $v = \eta \circ \gamma$ the element of the A-algebra C = Hom$_A$(E, B). For every element $u \in C$, $uv$ is the composite mapping

Then $uv = m \circ (u \otimes \eta) \circ h'' = u$. It is similarly proved that $vu = u$ and hence $v$ is unit element of C. Conversely, let the A-module A $\oplus$ E be given a unital algebra structure such that $(a, x)(a', x') = (aa', ax' + a'x)$ for $a, a'$ in A and $x, x'$ in E. Let B denote the A-algebra thus defined and let C be the A-algebra Hom$_A$(E, B). Suppose that C is unital and let $e:x \mapsto (\gamma(x), \lambda(x))$ be its unit element (where $\gamma(x) \in A$ and $\lambda(x) \in E$). On the other hand let $f$ be the element $x \mapsto (0, x)$ of C. An immediate calculation shows that $fe$ is the element

of C. The condition $fe = f$ implies the commutativity of the second diagram of (16) and it is similarly seen that the condition $ef = f$ implies the commutativity of the first diagram of (16).

A linear form $\gamma$ on E rendering diagrams (16) commutative is called a counit of the cogebra E. A cogebra admits at most one counit: for it is the unit element of the algebra Hom$_A$(E, A). A cogebra with a counit is called counital.