Proposition 8

Let $f$ be a mapping of A into B. If there exists a mapping $r$ (resp. $s$) of B into A such that $r \circ f$ (resp. $f \circ s$) is the identity mapping of A (resp. B), then $f$ is injective (resp. surjective). Conversely, if $f$ is surjective, there exists a mapping $s$ of B into A such that $f \circ s$ is the identity mapping of B. If $f$ is injective and if A $\neq \emptyset$, there exists a mapping $r$ of B into A such that $r \circ f$ is the identity mapping of A.

If there exists a mapping $r$ of B into A such that $r \circ f$ is the identity mapping of A, then the equality $f(x) = f(y)$, where $x \in A$ and $y \in A$, implies $x = r(f(x)) = r(f(y)) = y$, and so $f$ is injective. If there exists a mapping $s$ of B into A such that $f \circ s$ is the identity mapping of B, we have $B = f(s(B)) \subset f(A) \subset B$, so that $f$ is surjective. If $f$ is surjective, let T denote the term $\tau_y(y \in A \text{ and } f(y) = x)$. We have $f(T) = x$ for $x \in B$; if $s$ denotes the mapping $x \to T$ $(x \in B, T \in A)$, then $f \circ s$ is the identity mapping of B. Finally, suppose that $f$ is injective and that A $\neq \emptyset$, and let $a$ be an element of A. The relation

implies $(x, y) \in B \times A$ and therefore has a graph R with respect to the letters $x$, $y$. This graph is functional by reason of the hypothesis on $f$, and has B as its domain; and we have $R(x) = a$ if $x \in B - f(A)$, and $f(R(x)) = x$ if $x \in f(A)$. Hence the function $r = (R, B, A)$ is such that $r \circ f$ is the identity mapping of A.