Proposition 8

With the same notations as previously, suppose that $s_{\Phi}$ and $d_{\Phi}$ are bijective. Let $u$ and $\nu$ be isomorphisms of E onto E' and of F onto F' respectively. Then, in order that Φ be the inverse image of Φ' relative to u and v (that is to say that one have Φ(x, y) = Φ'(u(x), v(y)) whatever x ∈ E, y ∈ F may be), it is necessary and sufficient that one have u^{-1} = v^* and v^{-1} = u^*.

Indeed Φ'(u(x), v(y)) = Φ(x, y) may also be written Φ (x, u^*(v(y))) = Φ(x, y). If this holds whatever x ∈ E and y ∈ F may be, one has u^* ∘ v = 1 since Φ is non-degenerate. One therefore also has v^* ∘ u = 1 by (33). The converse is immediate.