Page 262 - HOW TO PROVE IT: A Structured Approach, Second Edition

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248 Functions
Commentary . Toprovethattwofunctionsareequal,weusuallyapplyTheorem
5.1.4. Thus, since f −1 ◦ f and i A are both functions from A to A, to prove that
they are equal we prove that ∀a ∈ A(( f −1 ◦ f )(a) = i A (a)).

Theorem 5.3.2 says that if f : A → B and f −1 : B → A, then each func-
tion undoes the effect of the other. For any a ∈ A, applying the function f
gives us f (a) ∈ B. According to Theorem 5.3.2, f −1 ( f (a)) = ( f −1 ◦ f )(a) =
i A (a) = a. Thus, applying f −1 to f (a) undoes the effect of applying f, giving
us back the original element a. Similarly, for any b ∈ B, applying f −1 we get
f −1 (b) ∈ A, and we can undo the effect of applying f −1 by applying f, since
f ( f −1 (b)) = b.
For example, let f : R → R be defined by the formula f (x) = 2x.You
should be able to check that f is one-to-one and onto, so f −1 : R → R, and
for any x ∈ R,

f −1 (x) = the unique y such that f (y) = x.
Because f −1 (x) is the unique solution for y in the equation f (y) = x,we
can find a formula for f −1 (x) by solving this equation for y. Filling in the
definition of f in the equation gives us 2y = x,so y = x/2. Thus, for every
x ∈ R, f −1 (x) = x/2. Notice that applying f to any number doubles the num-
ber and applying f −1 halves the number, and each of these operations undoes
the effect of the other. In other words, if you double a number and then halve
the result, you get back the number you started with. Similarly, halving any
number and then doubling the result gives you back the original number.
Are there other circumstances in which the composition of two functions
is equal to the identity function? Investigation of this question leads to the
following theorem.

Theorem 5.3.3. Suppose f : A → B.
1. If there is a function g : B → A such that g ◦ f = i A then f is one-to-one.
2. If there is a function g : B → A such that f ◦ g = i B then f is onto.
Proof

1. Suppose g : B → A and g ◦ f = i A . Let a 1 and a 2 be arbitrary elements of
A, and suppose that f (a 1 ) = f (a 2 ). Applying g to both sides of this equation
we get g( f (a 1 )) = g( f (a 2 )). But g( f (a 1 )) = (g ◦ f )(a 1 ) = i A (a 1 ) = a 1 ,
and similarly, g( f (a 2 )) = a 2 . Thus, we can conclude that a 1 = a 2 , and
therefore f is one-to-one.
2. See exercise 9.