Corollary

If $u$ is a bijection of $E'$ onto $E$ and $v$ is a bijection of $F$ onto $F'$, then $f \mapsto v \circ f \circ u$ is bijective.

Let $A$, $B$, $C$ be three sets and let $f$ be a mapping of $B \times C$ into $A$. For every $y \in C$ let $f(\mathord{\cdot}, y)$ be the partial mapping $x \mapsto f(x,y)$ of $B$ into $A$ (§ 3, no. 9); the function $y \mapsto f(\mathord{\cdot}, y)$ is a mapping of $C$ into $\mathfrak{F}(B, A)$. Conversely, for every mapping $g$ of $C$ into $\mathfrak{F}(B, A)$ there exists a unique mapping $f$ of $\mathrm{B} \times \mathrm{C}$ into A such that $g(y) = f(\bullet, y)$ for each $y \in \mathrm{C}$, namely the mapping $(x, y) \to (g(y))(x)$. Hence :