Math 201 — Homework 9
Due: 28 Oct. 2011
4.33 (a) If f : A → B and g : B → C are injections, then f(x) = f(y) implies x = y and
g(x) = g(y) implies x = y. Suppose that g(f(x)) = g(f(y)) (i.e., (g ◦ f)(x) =
(g ◦ f)(y)). Then since g is an injection we can conclude that f(x) = f(y), and
now since f is an injection we can conclude x = y. This shows that if (g◦f)(x) =
(g ◦ f)(y) then x = y which is equivalent to saying that g ◦ f is an injection.
(b) Suppose f : A → B and g : B → C are surjections, and consider a point z ∈ C.
Since g is a surjection there is some y ∈ B so that g(y) = z. Since f is a surjection
there is some x ∈ A so that f(x) = y. In particular we have (g◦f)(x) = g(f(x)) =
g(y) = z, showing that z is in the image of g ◦ f. But this is true for each z and
so g ◦ f is a surjection.
(c) If f and g are bijections then they are also injections so by part (a) their composition is an injection. Similarly, if f and g are bijections then they are also
surjections so by part (b) their composition is a surjection. Since the composition is an injection and a surjection then the composition is actually a bijection.
(d) To show that f−1 ◦ g−1 is (g ◦ f)−1 we consider the following:
f−1 ◦ g−1 ◦ g ◦ f = f−1 ◦ g−1 ◦ (g ◦ f)
= f−1 ◦ (g−1 ◦ g) ◦ f = f−1 ◦ id ◦ f = f−1 ◦ f = id,
where id is the identity map. Here we used the associative property of composition, along with the fact that the composition of the identity function with
any function κ gives κ. Similarly we have
g ◦ f ◦ f−1 ◦ g−1 = g ◦ f ◦ (f−1 ◦ g−1 )
= g ◦ (f ◦ f−1 ) ◦ g−1 = g ◦ id ◦ g−1 = g ◦ g−1 = id.
This shows that f−1 ◦ g−1 is an inverse of g ◦ f.
4.34 (a) This is true. We are given that h = g ◦ f is injective, and want to show that f
is injective. Since h is injective then if g(f(x)) = g(f(y)) (i.e., h(x) = h(y)) then
x = y. So now suppose that f(x) = f(y), then we have that g(f(x)) = g(f(y))
which implies x = y. This shows that f is injective.
(b) This is false. Consider the set A = C = {1} and B = {a, b} with f(1) = a and
g(a) = g(b) = 1. Then g ◦ f is the function which maps 1 to 1 but g is not a
surjection because it maps two values to 1.
(c) This is false. We can use the same example as (b), notice that the function g ◦ f
maps 1 to 1 and covers all of C, but f does not cover all of B so is not a surjection.
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(d) This is true. We are given that h = g ◦ f is a surjection, and want to show that g
is surjective. So select any element c ∈ C then since h is surjective then there is
some a ∈ A so that h(a) = c, but h(a) = g(f(a)) and f(a) ∈ B. Therefore f(a)
maps to c, so c has a pre-image for the function g. Since this is true for every
c ∈ C then the function g must be a surjection.
4.35 (a) The answer is no. Let A = {a, b} and B = {1} with f(a) = f(b) = 1 and g(1) = a.
Then f(g(1)) = f(a) = 1 and so the function satisfies f(g(y)) = y for all y ∈ B
But the function f is not a bijection because it is not injective (or more basically
the sets have different sizes).
(b) The answer is no. Let A = {1} and B = {a, b} with f(1) = a and g(a) = g(b) = 1.
Then g(f(1)) = g(a) = 1 and so the function satisfies g(f(x)) = x for all x ∈ A.
But f(g(b)) = f(1) = a 6= b.
4.36 (a) We have that f ◦ g is the identity, and the identity is a bijective function so in
particular is a surjective function. Therefore by problem 4.34(d) we know that
f must be surjective.
(b) We have that g ◦ f is the identity, and the identity is a bijective function so in
particular is an injective function. Therefore by problem 4.34(a) we know that
f must be injective.
4.37 By problem 4.34(a) since f ◦ f is injective then f must be injective.
4.49 We are given A1 , A2 , . . . a sequence of countable sets. In particular, we can label the
elements of these sets as ai,j which indicates that this is the jth element of the ith
set (here by the “jth” element we mean that in the bijection between Ai and N this
is the element that is paired with j). In particular, this gives a bijection between the
union of the Ai and the set N × N. But we know that N × N is countable, and so we
can conclude that the union must also be countable (i.e., a composition of bijections
is a bijection).
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