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Last edited 4/14/2009 10:23:00 AM

Inverse of functions

I have not yet written this article.  Meanwhile look at the Wikipedia article on function inverses.

The number 1/2 is the "multiplicative inverse" of the number 2 because their product is 1. A similar relationship can hold between functions, but because functional composition is not normally commutative, one has to specify which way the composite is taken.

Definition:inverse of a function

[ label: invdef] If F:A→B and G:B→A, then G is a left inverse to F, and F is a right inverse to G, if

If G is both a left and a right inverse to F, in other words if both Equation  [inveq] and

hold, then G is an inverse to F.

Usage

A function that has an inverse is said to be invertible.

Fact

[ label: I] t follows from the definition that if G is an inverse to F, then F is an inverse to G.

Fact

[ label: [ label: ]]  otherways The definition of inverse function can be expressed in other ways equivalent to Definition  [invdef] .

[ label: firstway]  G is the inverse of F if and only if for all aA, G(F(a))=a and for all bB, F(G(b))=b. (Both equations must hold.)

G is the inverse of F if and only if the following diagrams commute:

Example

Let F:{1,3,5}→{2,3,4} be the function that takes 1 to 3, 3 to 4 and 5 to 2. Then the function G:{2,3,4}→{1,3,5} that takes 2 to 5, 3 to 1 and 4 to 3 is the inverse of F. (And F is the inverse of G.)

Example

Example  [invexs] ( [threeinv] ) above says that the squaring function is a left inverse to the square root function: squaring the positive square root gives you what you started with. It is not the inverse, however: taking the square root of the square won't give you the number you started with if it is negative. On the other hand, the cubing function is the inverse of the cube root function.

A function can have more than one left inverse (Problem  [manyleftinv] ) but not more than one inverse:

Theorem: uniqinvUniqueness Theorem for InversesIf F:A→B has an inverse G:B→A, then G is the only inverse to F.

Proof: The proof uses the definition of inverse, Theorem  [compassoc] and Example  [idfax] : If H:B→A is another inverse of F, then

END PROOF

Usage

The fact that if a function has an inverse, it has only one, means that we can give the inverse a name: The inverse of F, if it exists, is denoted F\inverse.

Remark

The uniqueness theorem also means we have a rule of inference: Given F:A→B and G:B→A,

Exercise

[ label: whinve]  Which of the following functions have inverses? If it has one, give the inverse (by describing its graph or by a formula).

Exercise

F:{1,2,3,4}→{3,4,5,6}, with 1‖→3, 2‖→4, 3‖→6, and 4‖→5.

Exercise

G:{1,2,3,4}→{3,4,5,6,7}, with 1‖→3, 2‖→4, 3‖→6, and 4‖→5.

Exercise

H:{1,2,3,4}→{3,4,5,6} with 1‖→3, 2‖→5, 3‖→6, and 4‖→5.

Exercise

n‖→2n:N→N.

Exercise

n‖→n+1:N→N.

Exercise

n‖→n+1:Z→Z.

Exercise

n‖→(1/n):N-{0}→R.

Exercise

K:{1,2,3,4,5}→{1,2,3} with K(n)= floor ((n+1)/2).

Answer: Only (a) and (f) have inverses. For (a) the inverse is F\inverse:{3,4,5,6}→{1,2,3,4} with graph ({<3,1>,<4,2>,<6,3>,<5,4>)}. For (f) it is n&Verbar;→n-1:Z→Z.

Exercise

Which of the functions in Exercise  [whinve] have (a) left inverses, (b) right inverses? Answer: All except (c) and (h) have left inverses. (a), (f) and (h) have right inverses.

Exercise

Show that if a function G has an inverse F, then it has only one left inverse and that is F. Answer: If L is a left inverse of G:A→B, then for any x in the domain of G, L=Lø\idB =Lø(GøF)=(LøG)øF=\idA øF=F.

Exercise

[ label: manyleftinv]  Let I: R+ →R denote the inclusion function. Show that I has infinitely many left inverses.

The inverse of the composite of two functions which have inverses is the composite of the inverses, only in the reverse order:

Theorem: shoesockThe Shoe-Sock TheoremIf F:A→B and G:B→C both have inverses, then

Remark

The name comes from the fact that the inverse of putting on your socks and then putting on your shoes is taking off your shoes and then taking off your socks.

Proof: To prove the Shoe-Sock Theorem, we will prove that

and

and then apply Rule ( [invinf] ), page . To prove Equation  [shs1] , we note that the following diagram commutes: the left and right triangles are the diagrams in Figure ( [iddags] ), and the middle triangle is the left triangle in Figure ( [iddaga] ).

Equation  [shs2] is proved similarly. END PROOF

Another fact with a similar proof (left as an exercise) is:

Theorem: invinvIf F has an inverse, then (F\inverse)\inverse=F.

Remark

Another way of saying this is that a function is the inverse of its own inverse.

Exercise

Prove Theorem  [invinv] .

A final fact about inverses is the very important:

Theorem: invbijCharacterization of invertible functionsA function F:A→B has an inverse if and only if it is a bijection.

Remark

The importance of Theorem  [invbij] lies in the fact that having an inverse is defined in terms of functional composition but being a bijection is defined in terms of application of the function to an element of its domain. Any time a mathematical fact connects two such differently-described ideas it is probably useful.

Proof: I will go through the proof in some detail since it ties together several of the ideas of this chapter. We have to prove an equivalence, which means two implications.

First we show that if F has an inverse then it is a bijection. Suppose F has an inverse. We must show that it is injective and surjective. To show that it is injective, suppose F(a)=F(a'). Then

(The first and last equations follow from  [otherways] [firstway] and the middle equation from the substitution property, Theorem  [subprop2] .) So F is injective.

To show F is surjective, let bB. We must find an element aA for which F(a)=b. The element is F\inverse(b), since F(F\inverse(b))=b. Thus F is bijective.

Now we must show that if F is bijective, then it has an inverse. Suppose F is bijective. We must define a function G:B→A which is the inverse of F. Let bB. Then, since F is surjective, there is an element aA for which F(a)=b. Since F is injective there is exactly one such a. Let G(b)=a. Since F(a)=b, that says that G(F(a))=a, which is half of what we need to show to prove (using Definition  [invdef] ) that G=F\inverse. The other thing needed is that F(G(b))=b. But by definition of G, G(b) is the element a for which F(a)=b, so F(G(b))=b. That finishes the proof. END PROOF

Remarks

The second part of the proof says this: If F(a)=b, then F\inverse(b)=a, and if F\inverse(b)=a, then F(a)=b.

You might experiment with proving the contrapositives of the two implications in the preceding proof; some people find them easier to understand.

Exercise

Write a formula for the inverse of each of these bijections.

Exercise

x&Verbar;→ x2 : R+ → R+ .

Exercise

x&Verbar;→x-1:R→R.

Exercise

x&Verbar;→2x:R→R.

Exercise

x&Verbar;→(1/x):R→R.

Answer:

x&Verbar;→x.

x&Verbar;→x+1.

x&Verbar;→x/2.

This one is its own inverse.

Exercise

Prove that a function has a left inverse if and only if it is injective.

Exercise

Prove that a function has a right inverse if and only if it is surjective.

Exercise

Give a right inverse of the function GCD : N+ × N+ → N+ . (You are being asked to give the right inverse explicitly, not merely show it exists.) Answer: Let F:N→N×N be the function defined by F(n)=(n,n). Then GCD (F(n))= GCD (n,n)=n=\idN (n) so F is a right inverse of GCD . (There are many others.)

Exercise

Show that GCD : N+ × N+ → N+ does not have a left inverse. Answer: Suppose that GCD has a left inverse H:N→N×N. Then (1,2)=H( GCD (1,2))=H(1)=H( GCD (1,1