Corollary 3

*Suppose that $E$ is the direct sum of two free submodules $E_1$ and $E_2$. Let $Q_i$ be the restriction of $Q$ to $E_i$ and $p_i$ the canonical mapping of $C(Q_i)$ into $C(Q)$ ($i = 1, 2$). Then the linear mapping $p$ of $C(Q_1) \otimes C(Q_2)$ into $C(Q)$ deduced from the bilinear mapping $(a, b) \to p_1(a)p_2(b)$ of $C(Q_1) \times C(Q_2)$ into $C(Q)$ is a bijection.

It is enough in fact to consider the basis of $E$ obtained by taking the union of a basis of $E_1$ and a basis of $E_2$.*