Proposition 1

Let $E$ be a cogebra over $A$. In order that, for every associative $A$-algebra $B$, the $A$-algebra $\mathrm{Hom}_A(E, B)$ be associative, it is necessary and sufficient that the coproduct $c : E \to E \otimes_A E$ be such that the diagram

be commutative.

Let $B$ be an associative $A$-algebra and $u, v, w$ three elements of $C = \mathrm{Hom}_A(E, B)$. Let $m_3$ denote the $A$-linear mapping $B \otimes_A B \otimes_A B \to B$ which maps $b \otimes b' \otimes b''$ to $bb'b''$. By definition of the product on the algebra $C$, $(uv)w$ is the composite mapping

whilst $u(vw)$ is the composite mapping

It follows that if diagram (12) is commutative, the algebra $\mathrm{Hom}_A(E, B)$ is associative for every associative $A$-algebra $B$. To establish the converse, it suffices to show that there exists an associative $A$-algebra $B$ and three $A$-linear mappings $u, v, w$ of $E$ into $B$ such that the mapping $m_3 \circ (u \otimes v \otimes w)$ of $E \otimes E \otimes E$ into $B$ is injective. Take $B$ to be the $A$-algebra $\mathscr{T}(E)$ and $u, v, w$ the canonical mapping of $E$ into $\mathscr{T}(E)$. The mapping $m_3 \circ (u \otimes v \otimes w)$ is then the canonical mapping $E \otimes E \otimes E = \mathscr{T}^3(E) \to \mathscr{T}(E)$, which is injective.

When the cogebra $E$ satisfies the condition of Proposition 1, it is said to be coassociative.