This preview shows pages 1–2. Sign up to view the full content.
Homework #6 Solutions
1)
Let f : A
→
B and g: B
→
C denote two functions. If both f and g are surjective, prove
that the composition g
°
f: A
→
C is a surjection as well.
To prove g
f is surjective, we need to prove that for all elements c
∈
C, there exists
an element a such that
g
f (a) = c.
Let c be an arbitrary element from the set C.
Since g is surjective, we know there exists an element b such that g(b) = c.
Similarly, for this element b, since f is surjective, we know there exists an element a
such that f(a) = b.
Substituting, we have g(f(a)) = c.
Thus, we have shown that for an arbitrarily chosen element c, we can always find an
element a such that g
f (a) = c.
2)
Let g: A
→
A be a bijection. For n
≥
2, define g
n
= g
°
g
°
...
°
g, where g is composed
with itself n times. Prove that for n
≥
2, that (g
n
)
1
= (g
1
)
n
.
Use induction on n.
Base case: n=1. LHS = (g
1
)
1
= g
1
, the inverse of g.
RHS = (g
1
)
1
= g
1
, also the inverse of g.
Inductive hypothesis: Assume for an arbitrary value of n=k that (g
k
)
1
= (g
1
)
k
.
Inductive step: Under this assumption we must show that (g
k+1
)
1
= (g
1
)
k+1
.
(g
k+1
)
1
= (g
°
g
k
)
1
= (g
k
)
1
°
g
1
, by the rule for the inverse of a function composition.
= (g
1
)
k
°
g
1
, by the inductive hypothesis
= (g
1
)
k+1
, by the definition of function composition.
This proves the inductive step so we can conclude that (g
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
This document was uploaded on 07/14/2011.
 Spring '09

Click to edit the document details