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Last edited 4/14/2009 10:17:00 AM

COMPOSITION OF FUNCTIONS

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[ label: funccompsec] 

Definition:composition of functions

[ label: funccompdef] If F:A→B and G:B→C, then GøF:A→C is the function defined for all aA by (GøF)(a)=G(xxF(a)). GøF is the composite of F and G, and the operation " ø" is called functional composition.

How to think about composition

The composite of two functions is obtained by feeding the output of one into the input of the other. Suppose F:A→B and G:B→C are functions. If a is any element of A, then F(a) is an element of B, and so G(F(a)) is an element of C. Thus applying F, then G, gives a function from A to C, and that is the composite GøF:A→C.

Remarks

You may be familiar with the idea of functional composition in connection with the chain rule in calculus.

Our definition of GøF requires that the codomain of F be the domain of G. Actually, the expression G(F(a)) makes sense even if cod F is only included in dom G, and many authors allow the composite GøF to be formed in that case, too. We will not follow that practice here.

Example

If A={1,2,3,4}, B={3,4,5,6}, C={1,3,5,7}, F is defined by F(1)=F(3)=5, F(2)=3 and F(4)=6, and G is defined by G(3)=7, G(4)=5, G(5)=1 and G(6)=3, then GøF takes 1‖→1, 2‖→7, 3‖→1 and 4‖→3.

Warning

Applying the function GøF to an element of A involves applying F, then G - in other words, the notation " GøF" is read from right to left.

Functional composition is associative when it is defined:

Theorem: compassocIf F:A→B, G:B→C and H:C→D are all functions, then Hø(GøF) and (HøG)øF are both defined and

Proof: Let aA. Then by applying Definition  [funccompdef] twice,

and similarly

so Hø(GøF)=(HøG)øF. END PROOF

Warning

Commutativity is a different story. If F:A→B and G:B→C, GøF is defined, but FøG is not defined unless A=C. If A=C, then GøF:A→C and FøG:C→A, so normally FøG≠GøF. Commutativity may fail even when A=B=C: For example, let S=x‖→ x2 :R→R and T=x‖→x+1:R→R. Then for any xR, S(T(x))=(x+1 )2 and T(S(x))= x2 +1, so SøT≠TøS.

Pondering the following examples of functional composition may be helpful in understanding the idea of composition.

Example

[ label: invexs]  Let \SQ=x‖→ x2 :R→ R+ and \SQRT=x‖→x: R+ →R. ( x denotes the nonnegative square root of x.) Let \ABS denote the absolute value function from R to R. Then the following are true.

\SQRTø\SQ=\ABS:R→R.

[ label: twoinv]  (\SQRTø\SQ)\restr R+ =\id R+ .

[ label: threeinv]  \SQø\SQRT=\id R+ .

Example

[ label: idfax]  If F:A→B is any function, then

Fø\idA =F and

\idB øF=F.

This is analogous to the property that an identity element for a binary operation has (see  [idforbin] ), but in fact composition of functions is not a binary operation since it is not defined for all pairs of functions.

Example

If AB and BC, and i:A→B and j:B→C are the corresponding inclusion functions, then jøi is the inclusion of A into C.

Example

If F:A→B and CA with inclusion map i:C→A, then F\restrC=Føi. In other words, restriction is composition with inclusion.

Exercise

Describe explicitly (give the domain and codomain and either the graph or a formula) the composite GøF if

Exercise

F:{1,2,3,4}→{3,4,5,6}, with 1‖→3, 2‖→4, 3‖→ 6, and 4‖→5, and G:{3,4,5,6}→{1,3,5,7,9} with 3‖→1, 4‖→7, 5‖→7 and 6‖→3.

Exercise

F:x‖→ x3 :R→R, G:x‖→2x:R→R.

Exercise

F:x‖→2x:R→R. G:x‖→ x3 :R→R,

Exercise

F= inclusion :N→R, G:x‖→(x/2):R→R.

Exercise

F= p1 :R×R→R, G:x‖→(3,x):R→R×R.

Answer:

GøF:{1,2,3,4}→{1,3,5,7,9}, graph ({<1,1>,<2,7>,<3,3>,<4,7>)}.

GøF:R→R, (GøF)(x)=2 x3 .

GøF:R→R, (GøF)(x)=8 x3 .

n&Verbar;→(n/2):N→R.

<x,y>&Verbar;→<3,x>:R×R→R×R.

Exercise

[ label: compperm]  Let F:A→B, G:B→C. Show the following facts:

Exercise

If F and G are both injective, so is GøF.

Exercise

If F and G are both surjective, so is GøF.

Exercise

If F and G are both bijective, so is GøF.

Exercise

If GøF is surjective, so is G.

Exercise

If GøF is injective, so is F.

Exercise

Give examples of functions F and G for which GøF is defined and

Exercise

F is injective but GøF is not.

Exercise

G is surjective but GøF is not.

Exercise

GøF is injective but G is not.

Exercise

GøF is surjective but F is not.

Answer: (a) Let F:R→R be the identity function on R and G:R→R the squaring function.

(d) Let F: R+ →R be the inclusion function and G:R→ R+ the absolute value function.

Exercise

hard Let A, B and C be sets.

Exercise

Prove that if F:A→B is a function and C is nonempty, then G&Verbar;→FøG:\fsCA→\fsCA is a function which is injective if and only if F is injective, and surjective if and only if F is surjective.

Exercise

Prove that if H:B→C is a function and A has more than one element, then G&Verbar;→(GøH):\fsCA→\fsBA is a function which is injective if and only if H is surjective, and surjective if and only if H is injective.