Proposition 9

(a) Let E, F, G be sets, let $g$ be a mapping of E onto F and $f$ a mapping of E into G. Then there exists a mapping $h$ of F into G such that $f = h \circ g$ if and only if the relation $g(x) = g(y)$ (where $x \in \mathrm{E}$, $y \in \mathrm{E}$*) implies the relation* $f(x) = f(y)$. The mapping $h$ is then uniquely determined by $f$; if $s$ is a section of $g$, we have $h = f \circ s$.

(b) Let E, F, G be sets, let $g$ be an injection of F into E, and let $f$ be a mapping of G into E. Then there exists a mapping $h$ of G into F such that $f = g \circ h$ if and only if $f(\mathrm{G}) \subset f(\mathrm{F})$. The mapping $h$ is uniquely determined by $f$; if $r$ is a retraction of $g$, we have $h = r \circ f$.

(a) If $f = h \circ g$, the relation $g(x) = g(y)$ (where $x \in \mathrm{E}$, $y \in \mathrm{E}$) clearly implies $f(x) = f(y)$. And for every section $s$ of $g$ we have

which shows that $h$ is uniquely determined by $f$. Conversely, suppose that the relation $g(x) = g(y)$ implies $f(x) = f(y)$; let $s$ be a section of $g$, and let $h = f \circ s$; then for every $x \in E$ we have $g(s(g(x))) = g(x)$, hence $f(s(g(x))) = f(x)$, that is, $h(g(x)) = f(x)$ and therefore $f = h \circ g$.

(b) If $f = g \circ h$, then clearly $f(G) \subset f(F)$, and for every retraction $r$ of $g$ we have $h = (r \circ g) \circ h = r \circ f$, which shows that $h$ is uniquely determined by $f$. Conversely, suppose that $f(G) \subset f(F)$; let $r$ be a retraction of $g$, and put $h = r \circ f$; for every $x \in G$, there exists $y \in F$ such that $f(x) = g(y)$, so that

and therefore $f = g \circ h$.