Definition 3.3.10 (Composition). Let \(f:X\to Y\) and \(g:Y\to Z\) be two functions, such that the range of \(f\) is the same set as the domain of \(g\). We then define the composition \(g\circ f:X\to Z\) of two functions \(g\) and \(f\) to be the function defined explicitly by the formula $$ (g\circ f)(x):=g(f(x)) $$ If the range of \(f\) does not match the domain of \(g\), we leave the composition \(g\circ f\) undefined.
Definition 3.3.10
-
Lemma 3.3.12
\(=\) { let \(p=f\circ g\),
\(~~~\) Definition 3.3.10, \(p:Y\to W\) }\(=\) { let \(p=g\circ h\),
\(~~~\) Definition 3.3.10, \(p:X\to Z\) }\(=\) { let \(q=f\circ p\),
\(~~~\) Definition 3.3.10, \(q:X\to W\) }\(=\) { let \(q=p\circ h\),
\(~~~\) Definition 3.3.10, \(q:Z\to W\) }\(=\) { Definition 3.3.10 }
\(Proof~by~Tao\). Since \(f\circ g\) is a function from \(X\) to \(Z\), \(f\circ(g\circ h)\) is a function from \(X\) to \(W\). Simliarly \(f\circ g\) is a function from \(Y\) to \(W\), and hence \((f\circ g)\circ h\) is a function from \(X\) to \(W\). Thus \(f\circ(g\circ h)\) and \((f\circ g)\circ h\) have the same domain and range. In order to check that they are equal, we see from Definition 3.3.7 that we have to verify that \((f\circ(g\circ h))(x)=((f\circ g)\circ h)(x)\) for all \(x\in X\). But by Definition 3.3.10 $$ (f\circ (g\circ h))(x)=f((g\circ h)(x)) $$
-
Exercise 3.3.8
\(=\) { Definition 3.3.10 : function composition }
-
Exercise 3.3.5
\(\equiv\) { Definition 3.3.10 }
\(\equiv\) { Definition 3.3.10 }
-
Exercise 3.3.4
\(\equiv\) { Definition 3.3.10 }
-
Exercise 3.3.2
\(\equiv\) { Definition 3.3.10 }
-
Exercise 3.3.1
\(\equiv\) { Definition 3.3.10 }