Remark

In the proof of Proposition 10, a total ordering on $I$ was used to define an algebra structure on the tensor product $\bigotimes_{i \in I} E_i$ of the A-modules

E_i. If the ordering on I is changed, another multiplicative structure arises on $\bigotimes_{i \in I} E_i$, but the new algebra thus obtained is canonically isomorphic to the above, since both are solutions of the same universal mapping problem. For example, when $I = \{1, 2\}$, the canonical isomorphism of the algebra $E_1 \overset{\varepsilon}{\otimes}_A E_2$ onto the algebra $E_2 \overset{\varepsilon}{\otimes}_A E_1$ maps $x_1 \otimes x_2$ to $\varepsilon_{2,1}(\alpha, \beta)x_2 \otimes x_1$, where $x_1$ is homogeneous of degree $\alpha$ and $x_2$ homogeneous of degree $\beta$.

Let J be a subset of I and, for each $i \in J$, consider the canonical homomorphism $h_i : E_i \to \varepsilon \bigotimes_{i \in I} E_i = E$. By virtue of relations (25) a canonical homomorphism $h : E' = \varepsilon \bigotimes_{i \in J} E_i \to E$ is derived canonically (by Proposition 10) from these homomorphisms, such that, for all $i \in J$, $h'_i = h \circ h_i$, $h'_i$ being the canonical homomorphism $E_i \to E'$. Taking the total ordering on J induced by that chosen on I, we obtain

where the middle term is the product of the ordered sequence $(h_i(x_i))_{i \in J}$ and where, in the right hand term, $x'_i = x_i$ for $i \in J$, $x'_i = e_i$ for $i \notin J$.