Proposition 6

If f is a mapping of A into B, and g is a mapping of B into C, then $g \circ f$ is a mapping of A into C.

Let F, G be the graphs of $f$, $g$, respectively, and let us show that $\mathrm{G} \circ \mathrm{F}$ is a functional graph. Let $x$, $z$, $z'$ be objects such that $(x, z) \in \mathrm{G} \circ \mathrm{F}$, $(x, z') \in \mathrm{G} \circ \mathrm{F}$. There exist objects $y$, $y'$ such that

Since F is a functional graph, we have $y = y'$ and hence $(y, z') \in \mathrm{G}$. Since G is a functional graph, it follows that $z = z'$, which proves our assertion. Furthermore, the domain of $g \circ f$ is evidently A, and the proof is complete.

¶ The function $g \circ f$ may also be written as $x \to g(f(x))$, or as $gf$ when there is no risk of confusion.