Proposition 2

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

(where $\sigma$ is the symmetry homomorphism such that $\sigma(x \otimes y) = y \otimes x$) is commutative (in other words, it suffices that the cogebras E be identical with its opposite (no. 1, Example 2).

Let B be a commutative A-algebra and $u, v$ two elements of $C = \mathrm{Hom}_A(E, B)$.

By definition of the product in C, $uv$ and $vu$ are respectively equal to the composite mappings

and

It follows that if diagram (15) is commutative the algebra $\mathrm{Hom}_A(E, B)$ is commutative for every commutative A-algebra B. To establish the converse, it suffices to show that there exist a commutative A-algebra B and two A-linear mappings $u, v$ of E into B such that $m \circ (u \otimes v): E \otimes E \to B$ is injective. Take B to be the algebra $S(E \oplus E)$ and $u$ (resp. $v$) the composition of the canonical mapping $E \oplus E \to S(E \oplus E)$ and the mapping $x \mapsto (x, 0)$ (resp. $x \mapsto (0, x)$) of E into $E \oplus E$. If $h: S(E) \otimes S(E) \to S(E \otimes E)$ is the canonical isomorphism (§ 6, no. 6, Proposition 9) and $\lambda: E \to S(E)$ is the canonical mapping, then $h^{-1} \circ m \circ (u \otimes v) = \lambda \otimes \lambda$. Now $\lambda \otimes \lambda$ is injective, for $\lambda(E)$ is a direct factor of $S(E)$ (II, § 3, no. 7, Corollary 5 to Proposition 7).

When the cogebra E satisfies the condition of Proposition 2, it is said to be cocommutative.