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Section 5.3 Composition and Inverse Functions
Definition 5.9 .
Let \(A\text{,}\) \(B\) and \(C\) be sets, and let \(f : A \to B\) and \(g : B \to C\) be functions. The composition of \(f\) and \(g\) is the function \(g \circ f : A \to C\) defined by
\begin{equation*}
(g \circ f )(x) = g( f (x))
\end{equation*}
for all \(x\in A\text{.}\)
Theorem 5.10 .
Let \(A, B, C\) and \(D\) be sets, and let \(f : A \to B\) and \(g : B \to C\) and \(h : C \to D\) be functions.
\((h \circ g) \circ f = h \circ (g \circ f )\) (Associative Law).
\(f \circ 1A = f and 1B \circ f = f\) (Identity Law).
Definition 5.11 .
Let \(A\) and \(B\) be sets, and let \(f : A \to B\) and g : \(B \to A\) be functions.
The function \(g\) is a right inverse for \(f\) if \(f \circ g = 1B\text{.}\)
The function \(g\) is a left inverse for \(f\) if \(g \circ f = 1A\text{.}\)
The function \(g\) is an inverse for \(f\) if it is both a right inverse and a left inverse.
Theorem 5.12 .
Let \(A\) and \(B\) be sets, and let \(f : A \to B\) be a function.
If \(f\) has an inverse, then the inverse is unique.
If \(f\) has a right inverse \(g\) and a left inverse \(h\text{,}\) then \(g = h\text{,}\) and hence \(f\) has an inverse.
If \(g\) is an inverse of \(f\text{,}\) then \(f\) is an inverse of \(g\text{.}\)
Exploration 5.14 .
Let \(A\) and \(B\) be sets, let \(U \subseteq A\) and \(V \subseteq C\) be subsets, and let \(f : A \to B\) and \(g : B \to C\) be functions. Prove that
\begin{equation*}
(g \circ f )(U) = g( f (U)) and (g \circ f )-1(V ) = f\inv (g-1(V ))\text{.}
\end{equation*}
