CHAPTER 1
1. α is neither 1-1 nor onto. β is 1-1 and onto.
α(A) = {1, 2},
β(B) = {1, 2, 3}, α(A ∩ B) = {1}, β(A ∪ B) = T.
2. β is onto. α and γ are 1-1.
α(N) = 2N, β(N) = N\{1}, γ(N) = {n3 : n ∈ N}.
3. 44 . 4! 4!.
4. (i) there is no x ∈ S such that α(x) = y.
(ii) there exist x1 , x2 ∈ S with x1 6= x2 such that α(x1 ) 6= α(x2 ).
(iii) there exists at least one x ∈ S such that α(x) 6= β(x).
5. 1. nm
2. n! if n ≤ m and 0 if n > m.
6. (a) each horizontal line intersects the graph of f at one point or
no intersection.
(b)each horizontal line intersects the graph of f at exactly one point.
7.
1 − 1 : a, b, c, e
onto : a, b, c.
8. a = ±1. a 6= 0.
9.
10.
Proof. For any s ∈ S, we have
(α ◦ ιS )(s) = α[ιS (s)] = α(s),
which implies
α ◦ ιS = α.
Similarly for ιS ◦ α = α.
11.β −1 (n) = n − 1.
12. (a ) f −1 (x) = x/2.
(b) f −1 (x) = x + 4.
(c) f −1 (x) = (x + 4)/2.
1
(d) f −1 (x) = x 3 .
1
2
CHAPTER 1
13.
Proof. Let α be an invertible mapping from S to T . Then
α ◦ α−1 = ιT ,
α−1 ◦ α = ιS .
which implies α−1 is also invertible and
[α−1 ]−1 = α.
14.
Proof. (i) Clearly,
α(A), α(B) ⊂ α(A ∪ B),
which implies
α(A) ∪ α(B) ⊂ α(A ∪ B).
Conversely, for any s ∈ A ∪ B, s ∈ A or s ∈ B. Thus, α(s) ∈ α(A) or
α(s) ∈ α(B), which implies
α(s) ∈ α(A) ∪ α(B).
Hence
α(A ∪ B) ⊂ α(A) ∪ α(B)
and the proof is concluded.
(ii) For any s ∈ A ∩ B, then s ∈ A and s ∈ B. Thus,
α(s) ∈ α(A) and α(s) ∈ α(B)
which implies
α(A ∩ B) ⊆ α(A) ∩ α(B).
Let S = {1, 2}, A = {1}, B = {2}, T = {a} and let α(1) = α(2) = a.
Then
α(A ∩ B) = ∅ ( {a} = α(A) ∩ α(B).
15.
Proof. Sufficiency: Suppose that α(A∩B) = α(A)∩α(B) for all A, B ⊂
S. For any s1 , s2 ∈ S with s1 6= s2 , we take A = {s1 } and B = {s2 }.
Then A ∩ B = ∅ and thus α(A ∩ B) = ∅. By our assumption,
α(A) ∩ α(B) = ∅.
That means
α(s1 ) 6= α(s2 )
and therefore α is 1 − 1.
CHAPTER 1
3
Necessity: Assume that α is 1 − 1. Then for any A, B ⊂ S and for
any t ∈ α(A) ∩ α(B),
t ∈ α(A) and t ∈ α(B).
Thus, there exist s1 ∈ A and s2 ∈ B such that
α(s1 ) = α(s2 ) = t.
Since α is 1 − 1, s1 and s2 must be the same element in S and therefore
s1 = s2 ∈ A ∩ B. Hence t ∈ α(A ∩ B) and the proof is concluded. 16 & 17. Example: Let S = {a}, T = {1, 2} and U = {x} and let
α(a) = 1, β(1) = β(2) = x.
18.
Proof. For any t ∈ T , since α is onto, there exists at least one s ∈ S
such that α(s) = t. By our assumption,
[β ◦ α](s) = [γ ◦ α](s),
which is equivalent to
β(t) = γ(t).
Hence β = γ.
19. Similar to 18.
20. Example: S = {a}, T = {1, 2}, U = {x, y} and let α(a) = 1,
β(1) = β(2) = γ(1) = x, γ(2) = y.
21. Similar to 20.
22. or not onto.
23.
Proof. (a) Since α and β are invertible, we have
α ◦ α−1 = ιT
α−1 ◦ α = ιS
β ◦ β −1 = ιU
β −1 ◦ β = ιT .
and
Therefore,
(β ◦ α) ◦ (α−1 ◦ β −1 ) = β ◦ (α ◦ α−1 ) ◦ β −1
= β ◦ ιT ◦ β −1
= β ◦ β −1
= ιU
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CHAPTER 1
and similarly
(α−1 ◦ β −1 ) ◦ (β ◦ α) = ιS .
Hence β ◦ α is invertible and its inverse is α−1 ◦ β −1 .
(b) β ◦ α is invertible implies that β ◦ α is both 1 − 1 and onto. Hence
α is 1 − 1 and β is onto.
24. (a) (c) (d) (e) (g) define operation on integers, where (c) (g) are
associative, (a) (d) (g) are commutative, none has identity.
25.
Proof. Assume, on the contrary, that m is an identity. Then
mn + 1 = n ∀n ∈ Z.
That is,
m=
n−1
∈
/ Z.
n
26. No, because 2−1 ∈
/ Z.
27. (a) (c)
28. (a) for some a, b, c ∈ S.
(b) for some a, b ∈ S.
(c) a ∗ e = e ∗ a = a doesn’t hold for some a ∈ S.
(d) a ∗ e = e ∗ a = a doesn’t hold.
a
a a
29. (a) b b
u u
v v
(b) No. If so,
(c) see (b).
b u v
b u v
a v u
v a b
u b a
then u = u ∗ v = v.
30. (a) 2Z = {2n : n ∈ Z}.
(b) 2Z = {2n : n ∈ Z}.
π
31. (a) ρ
σ
τ
(b) ρ
(c) No
π
π
π
τ
τ
ρ
π
ρ
σ
τ
σ
π
σ
σ
τ
τ
π
τ
π
τ
CHAPTER 1
5
(d)ρ
(e) Yes.
32. (a)
Proof. Since α1,0 (x) = x for every x ∈ A, α1,0 = ιS .
(b) If αa,b (x) = αa,b (y), then
ax + b = ay + b
which further implies x = y. Hence αa,b is one-to-one. Additionally,
for any z ∈ R, take x = z−b
∈ R. Then αa,b (x) = z and hence αa,b is
a
onto.
(c) (c, d) = ( a1 , − db ).
33. (a) Reduced x to a times vector ~x
(b) Reduced x to a times vector ~x in the opposite direction
(c) Translate b to the right
34. (a)
Proof. For any αx,0 , αy,0 ∈ B, we have
[αx,0 ◦ αy,0 ](z) = y(xz) = (yx)z = αxy,0 (z) for every z ∈ R,
which shows
αx,0 ◦ αy,0 = αxy,0 ∈ B.
Hence B is closed with respect to ◦ and ◦ is an operation on B. Moreover, ◦ is associative, commutative and has identity α1,0 .
(b) It is easy to verify
α1,b ◦ α1,c = α1,b+c ∈ C.
That means ◦ is an operation on C and it is commutative, associative
and has identity α1,0 .
(c)
αa,b = α1,b ◦ αa,0 .
35. D = α1,Z = {α1,n : n ∈ Z}.
36.
Proof. Let α : S → T be an invertible mapping and let β, γ be inverses
of α. Then
β = β ◦ ι =β ◦ (α ◦ γ)
=(β ◦ α) ◦ γ
=ι ◦ γ
=γ.
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CHAPTER 1
37.
Proof.
[γ ◦ (β ◦ α)](s) = γ[(β ◦ α)(s)] = γ[β(αs)]
= [γ ◦ β](αs)] = [(γ ◦ β) ◦ α](s) for any s.
That implies the desired result.
38. Yes.
39.
Proof. Let S and T be linear transformations on V . For any u, v ∈ V
and s, t ∈ R,
(S ◦ T )(su + tv) =S[T (su + tv)]
=S[sT (u) + tT (v)]
=sS[T (u)] + tS[T (v)]
=s[S ◦ T ](u) + t[S ◦ T ](v) ∈ V,
which yields that S ◦ T is a linear transformation on V . Hence composition is an operation on V .