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

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

[ label:
funccompsec]

[ 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.

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.

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.

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.

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

Hø(GøF)=(HøG)øF |

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

((Hø(GøF) )(a)=H
((GøF)(a) )=H(G(F(a))) |

and similarly

(((HøG)øF
)(a)=(HøG)(F(a))=H(G(F(a))) |

so Hø(GøF)=(HøG)øF. END PROOF

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.

[ 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+ .

[ 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.

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.

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

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

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.

F:x‖→ x3
:R→R, G:x‖→2x:R→R.

F:x‖→2x:R→R.
G:x‖→ x3 :R→R,

F= inclusion :N→R,
G:x‖→(x/2):R→R.

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‖→(n/2):N→R.

<x,y>‖→<3,x>:R×R→R×R.

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

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

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

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

If GøF is surjective, so is G.

If GøF is injective, so is F.

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

F is injective but GøF is not.

G is surjective but GøF is not.

GøF is injective but G is not.

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.

hard Let A, B and C
be sets.

Prove that if F:A→B
is a function and C is
nonempty, then G‖→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.

Prove that if H:B→C
is a function and A has
more than one element, then G‖→(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.