Definition 3.3.7 (Equality of functions). Two functions \(f:X\to Y,~g:X\to Y\) with the same domain and range are said to be equal, \(f=g\) if and only if \(f(x)=g(x)\) for all \(x\in X\). (If \(f(x)\) and \(g(x)\) agree for some values of \(x\), but not others, then we do not consider \(f\) and \(g\) to be equal 1 .)
Remark 3.3.10.* It is not immediately apparent that Definition 3.3.7 is compatible with the axioms of eqaulity in Addendix A.7, although Exercise 3.3.1 below provides evidence towards this compatibility. There are at least three ways to address this issue. One is to regard Definition 3.3.7 as an axiom about equality of functions rather than a definition. Another is to provide a more explicit definition of a function in which Definition 3.3.7 becomes a theorem; for instance, one can define a function \(f:X\to Y\) to be an ordered triple \((X,Y.Z)\) consisting of a domain set \(X\), a range set \(Y\), and a graph \(G=\{(x,f(x)):x\in X\}\) that obeys the vertical line test, and use this latter graph to define the value of \(f(x)\in Y\) for each element \(x\) of the domain (cf. Exerciese 3.5.10). A third way is to start with a mathematical universe \(\mathcal{U}\) without any funtions in it, and use Definition 3.3.7 to create a larger extension of this universe that contains function objects that behave as specified as in Definition 3.3.7. This final procedure however requires a bit more of the formalism of logic and model theory than is provided by this text, and so will not be detailed here.
\(Proof~by~Tao\). 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 $$ (f\circ (g\circ h))(x)=f((g\circ h)(x)) $$
\(\equiv\) { Claim 1, Claim2, Claim 3, and Definition 3.3.7, modus ponens}
\(\equiv\) { Definition 3.3.7 }
\(\equiv\) { Definition 3.3.7 }
Exercise 3.3.1. Show that the definition of equality in Definition 3.3.7 is reflexive, symmetric, and transitive. Also verity the substitution property: if \(f,\tilde{f}:X\to Y\) and \(g,\tilde{g}:Y\to Z\) are functions such that \(f=\tilde{f}\) and \(g=\tilde{g}\), then \(g\circ f=\tilde{g}\circ\tilde{f}\).
Exercise 3.3.1. Show that the definition of equality in Definition 3.3.7 is reflexive, symmetric, and transitive. Also verity the substitution property: if \(f,\tilde{f}:X\to Y\) and \(g,\tilde{g}:Y\to Z\) are functions such that \(f=\tilde{f}\) and \(g=\tilde{g}\), then \(g\circ f=\tilde{g}\circ\tilde{f}\).