Natalie Lowe
Chris Kelly
Math 403 Projecl
HoMEwoRx
6 SoLUTloNs:
1) Prove that if f: A--+B is a bijection, thon there exists a bijection g: B + A such
thrt g o f = 1 and f o g = I (Nherc | denobs,afidentq.
+^4
P-regt
First,lets
assume we have some x e Aandy e B such that f(x) = y.
Since f is a bijectiorq then it is a one-to-one and onto by definition, so
f(x)
=
f(x')
+ x:x'
and V y e B, I anx € A such that f(x) = y.
Then.
suDpose we havea bijection
g.
such thal
g(y) x. tfe
Ifwe take the cornposition
ofthe two bijections, then .{:f-lr,
g(f(x)) = g(y) = x, because
g(y) = x and
y = f(x)
and, f(g(y)): f(x) = y,
because f(x) - y and x = g(y). .cr-ai
4^@ -r .t*Ft"..
\A< .,,42d f. F ro .-<
\14 t4 € av,-z-
So,
g(f(x))
-x r: x andf(e8)) = y t= y.
. . The composition
oftlrc two bijections is the identity.
2) Ex. Z.lt Proye that the compositiotr oI two bijections is a bijection.
fueef
Letf:A+B and
g:B+C be two bijections.
Letx € A,y e B,andz e C be such that f(x)
= y and
g(y)
: z.
Ald, by definition, fand g are
bijections, and ole-to-one and onto.
Since we know that f and
g are one-to-one, then
we know tlnt there is a rmique x, such
that f(x) = y and
also a unique
y, such that f(y) = z.
Now
C(y)
=
C(y')
=>
y=y'orinotherwordsz=z' - y=y'
and,
f(x) = f(x') = x = x' or
in
odrer words
y = y' + x=x'.
Thus,
g(f(x))
= g(f(x')) = x=x'
or in
other words
g(y)
=
g(y')
:+ x=x!or z=z' > x=x'.
So, there is a urique x € A such that
g(f(x)): z. And, the composition
g o fis otre-to-
one. Since
g and fare onto,
each z e Chasay e B such
that
g(y): z, and each
