1.3.4: Inverses of Functions
We think of functions as maps. An obvious question to ask about functions, then, is whether the mapping can be “reversed.” For instance, the successor function \(f(x) = x + 1\) can be reversed, in the sense that the function \(g(y) = y - 1\) “undoes” what \(f\) does.
But we must be careful. Although the definition of \(g\) defines a function \(\Int \to \Int\) , it does not define a function \(\Nat \to \Nat\) , since \(g(0) \notin \Nat\) . So even in simple cases, it is not quite obvious whether a function can be reversed; it may depend on the domain and codomain.
This is made more precise by the notion of an inverse of a function.
A function \(g \colon B \to A\) is an inverse of a function \(f \colon A \to B\) if \(f(g(y)) = y\) and \(g(f(x)) = x\) for all \(x \in A\) and \(y \in B\) .
If \(f\) has an inverse \(g\) , we often write \(f^{-1}\) instead of \(g\) .
Now we will determine when functions have inverses. A good candidate for an inverse of \(f\colon A \to B\) is \(g\colon B \to A\) “defined by” \[g(y) = \text{“the” $x$ such that $f(x) = y$.}\nonumber\] But the scare quotes around “defined by” (and “the”) suggest that this is not a definition. At least, it will not always work, with complete generality. For, in order for this definition to specify a function, there has to be one and only one \(x\) such that \(f(x) = y\) —the output of \(g\) has to be uniquely specified. Moreover, it has to be specified for every \(y \in B\) . If there are \(x_1\) and \(x_2 \in A\) with \(x_1 \neq x_2\) but \(f(x_1) = f(x_2)\) , then \(g(y)\) would not be uniquely specified for \(y = f(x_1) = f(x_2)\) . And if there is no \(x\) at all such that \(f(x) = y\) , then \(g(y)\) is not specified at all. In other words, for \(g\) to be defined, \(f\) must be both injective and surjective.
Every bijection has a unique inverse.
Proof. Exercise. ◻
Prove Proposition \(\PageIndex{1}\) . That is, show that if \(f\colon A \to B\) is bijective, an inverse \(g\) of \(f\) exists. You have to define such a \(g\) , show that it is a function, and show that it is an inverse of \(f\) , i.e., \(f(g(y)) = y\) and \(g(f(x)) = x\) for all \(x \in A\) and \(y \in B\) .
However, there is a slightly more general way to extract inverses. We saw in section 3.2 that every function \(f\) induces a surjection \(f' \colon A \to \ran{f}\) by letting \(f'(x) = f(x)\) for all \(x \in A\) . Clearly, if \(f\) is an injection, then \(f'\) is a bijection, so that it has a unique inverse by Proposition \(\PageIndex{1}\). By a very minor abuse of notation, we sometimes call the inverse of \(f'\) simply “the inverse of \(f\) .”
Show that if \(f\colon A \to B\) has an inverse \(g\) , then \(f\) is bijective.
Every function \(f\) has at most one inverse.
Proof. Exercise. ◻
Prove Proposition \(\PageIndex{2}\) . That is, show that if \(g\colon B \to A\) and \(g'\colon B \to A\) are inverses of \(f\colon A \to B\) , then \(g = g'\) , i.e., for all \(y \in B\) , \(g(y) = g'(y)\) .