Example

(1) Take E to be the graded cogebrà $T(M)$ (no. 1) and suppose that N, $N'$, $N''$ have the trivial graduation. A graded homomorphism of degree $-p$ of $T(M)$ into N (resp. $N'$, $N''$) then corresponds to a multilinear mapping of $M^p$ into N (resp. $N'$, $N''$). Given a multilinear mapping $u : M^p \to N$ and a multilinear mapping $v : M^q \to N'$, the above method allows us to deduce a multilinear mapping $m(u, v) : M^{p+q} \to N''$ called the product (relative to $m$) of $u$ and $v$. Formulae (3) (no. 1) and (11) show that, for $x_1, \ldots, x_{p+q}$ in M,

(2) Take E to be the graded cogebrà $S(M)$ (no. 1), preserving the same hypotheses on N, $N'$, $N''$. A graded homomorphism of degree $-p$ of $S(M)$ into N then corresponds to a symmetric multilinear mapping of $M^p$ into N (§ 6, no. 3). Then we derive from a symmetric multilinear mapping $u : M^p \to N$ and a symmetric multilinear mapping $v : M^q \to N'$ a symmetric multilinear mapping $m(u, v) : M^{p+q} \to N''$, also denoted (to avoid confusion) by $u._m v$ (or even $u.v$) and called the symmetric product (relative to $m$) of $u$ and $v$. Formulae (6) (no. 1) and (11) show that, for $x_1, \ldots, x_{p+q}$ in M,

the summation being taken over permutations $\sigma \in \mathfrak{S}_{p+q}$ which are increasing on each of the intervals $[1, p]$ and $[p+1, p+q]$.

(3) Take E to be the graded cogebra $\bigwedge(M)$ (no. 1). Then we deduce similarly from an alternating multilinear mapping $u : M^p \to N$ and an alternating multilinear mapping $v : M^q \to N'$ an alternating multilinear mapping $m(u, v) : M^{p+q} \to N''$, also denoted by $u \wedge_m v$ or $u \wedge v$ and called the alternating product (relative to $m$) of $u$ and $v$. Formulae (9) (no. 1) and (11) show in this case that, for $x_1, \ldots, x_{p+q}$ in $M$,

the summation again being taken over permutations $\sigma \in \mathfrak{S}_{p+q}$ which are increasing on each of the intervals $[1, p]$ and $[p+1, p+q]$.

We return to the case where E is an arbitrary graded cogebra (of type N) and assume that the three modules N, N', N'' are all equal to the underlying A-module of a graded A-algebra B of type $\mathbf{Z}$, the mapping $m$ being multiplication in B, so that $\tilde{m} : B \otimes_A B \to B$ is a graded A-linear mapping of degree 0. Thus a graded A-algebra structure is obtained on the graded A-module $\mathrm{Homgr}_A(E, B) = C$.

In particular, suppose that $B = A$ (with the trivial graduation), so that $\mathrm{Homgr}_A(E, A)$ is the graded dual $E^{*gr}$, which thus has a graded A-algebra structure.

Let F be another graded cogebra $c'$ its coproduct and $\phi : E \to F$ a graded cogebra morphism (no. 1); then the canonical graded morphism

is a graded algebra homomorphism. For $u, v$ in $\mathrm{Homgr}_A(F, B)$ and $x \in E$,

But by hypothesis $c'(\phi(x)) = (\phi \otimes \phi)(c(x))$, hence

and therefore $\tilde{\phi}(uv) = \tilde{\phi}(u)\tilde{\phi}(v)$, which proves our assertion.

In particular, the graded transpose $^t\phi : F^{*gr} \to E^{*gr}$ is a graded algebra homomorphism.