64
(c)
(f
0
g)1
=
{(I,
4), (2,3), (3,2),
(4,
1),
(5,
5n
g1
0
f
1
=
{(I,
4), (2,3), (3,2),
(4,
1),
(5,
5n
=
(f
0
g)I.
f
1
0
g1
=
{(I,
5), (2,3), (3,2), (4,4),
(5,
In
=I
(f
0
g)I.
Solutions
to
Exercises
20.
(a)[BB]ForxEB,(fog)(x)=fC:~I)=
2~
=
2t
)=22
x
=x.
xI

2
2x

2
xI
2(~)
2x
2x
(b) [BB] For
x
E
A,
(g
0
f)(x)
=
g(x:'2)
=
~1
=
x
_
(x
_
2)
=
2"
=
x
and so, by
x2
Proposition
3.2.7,
f
and
9
are inverses.
21.
(a) Suppose
0
E
A
and let a
=
f(O).
Then
0
=
fl(a)
=
fta)
which is not possible for any real
number
f(a).
(b)
Since
f
and
f4
each have domain
A,
we have only to prove that
f4(a)
=
a for all a
E
A.
Let
then a
be
some element
of
A.
Let
al
=
f(a),
a2
=
f(al)
=
f2(a),
a3
=
f(a2)
=
f3(a)
and
a4
=
f(a3)
=
f4(a).
We must show that
a4=
a. From
al
=
f(a),
we have a
=
f
1
(al) =
f(~J)
and so a2
=
f(al)
=
~.
From a3
=
f(a2),
we obtain a2
=
f
1
(a3)
=
f(~3)
and so
a4
=
f(a3)
=
.!..
=
_1_
==
a
as
desired.
a2
l/a
22.
(a) [BB] Suppose
g(b
1)
=
g(b2)
for
b1
,b2
E
B.
Since
f
is onto,
b
1
=
f(aJ)
and
b2
=
f(a2)
for
some alo a2
E
A.
Thus,
g(f(al))
=
g(f(a2));
that is,
9
0
f(al)
=
go
f(a2).
Since
9
0
f
is
onetoone,
al
=
a2.
Therefore,
b
1
=
f(al)
=
f(a2)
=
b2
proving that
9
is onetoone.
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 Summer '10
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 Logic, Graph Theory, Existential quantification, Zagreb, Zagreb bypass, nd

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