6.4 Composing functions

Let’s check explicitly that $g\circ f$ really does fit the formal definition of a function from Definition 6.1. We need to define a graph $H$ such that:

1. (i)
   ​

   for every $x\in A$, there exists $y\in C$ such that $(x,y)\in H$; and

2. (ii)
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   if $(x,y)\in H$ and $(x,z)\in H$ then $y=z$.

Let

be the graphs of the functions $f\colon A\to B$ and $g\colon B\to C$. Define the graph of the composition by

Note that this means, for every $(x,z)\in H$, there exists some $y\in B$ such that $(x,y)\in F$ and $(y,z)\in G$. Hence, by definition of $F$ and $G$, $y=f(x)$ and $z=g(y)=g(f(x))=g\circ f(x)$.

We now check the two criteria for our graph $H$ from Definition 6.1.

1. (i)
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   Existence. Fix $x\in A$. Since $F$ is the graph of the function $f$, there exists $y\in B$ with $(x,y)\in F$. Because $G$ is the graph of the function $g$ and its domain is all of $B$, for that $y$ there exists $z\in C$ with $(y,z)\in G$. Thus $(x,z)\in H$. Hence for every $x\in A$ there is some $z\in C$ with $(x,z)\in H$.

2. (ii)
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   Uniqueness. Suppose $(x,z)\in H$ and $(x,w)\in H$. Then there exist $y,y^{\prime}\in B$ with $(x,y)\in F$, $(y,z)\in G$ and $(x,y^{\prime})\in F$, $(y^{\prime},w)\in G$. Because $F$ is the graph of a function, $(x,y)$ and $(x,y^{\prime})$ force $y=y^{\prime}$. Now since $G$ is the graph of a function and $(y,z),(y,w)\in G$, we obtain $z=w$, as required.

Therefore $H$ satisfies the existence requirement (i) and the uniqueness requirement (ii), so it is indeed the graph of a function from $A$ to $C$. This function is precisely the composition $g\circ f$, given by $(g\circ f)(x)=g(f(x))$.