SURJECTIVITY, INJECTIVITY AND BIJECTIVITY
5 minute review. Recap the definitions of surjectivity, injectivity and bijectivity,
drawing pictures if appropriate.
Class warm-up. Work through a few parts from Question 1 below (e.g. (a)-(c)).
Problems. Choose from the below.
1. For each function below, is it (i) surjective, (ii) injective, (ii) bijective? If you
have answered no to (i) or (ii), give a clear counter-example to back up your
claim. If you have answered yes to (iii), state the inverse function.
(a) f : R → R,
f (a) = sin a;
(b) f : R → [−1, 1], f (a) = sin a;
(c) f : [−π/2, π/2] → [−1, 1], f (a) = sin a;
(d) f : R → R,
f (a) = a4 ;
(e) f : R≥0 → R, f (a) = a4 ;
(f) f : R → R≥0 , f (a) = a4 ;
(g) f : R≥0 → R≥0 , f (a) = a4 ;
(h) f : Q≥0 → Q≥0 , f (a) = a2 ;
(i) f : Q\{0} → Q\{0}, f (a) = a1 ;
(j) f : C → R,
f (a) = |a|;
(k) f : C → C,
f (a) = a.
2. Let f : Q → Q be given by f (a) =
surjective?
a3 + 1
. Show that f is injective. Is f
a3 − 2
a−3
. Show that f is injective. Can
a−2
you find a ∈ R\{2} such that f (a) = 1? Is f bijective? If not, can you modify
the codomain to make a bijective function (and prove that it is bijective)?
3. Let f : R\{2} → R be given by f (a) =
4. (a) Let f : (1, ∞) → R be given by f (x) =
x+1
. Show that f is injective.
x2 + 1
(b) What is the smallest value of a for which we can conclude that the
function f : (a, ∞) → R with the same rule as above is injective?
Homework: Chapter 1, Q9 from the problem booklet
1
2
SURJECTIVITY, INJECTIVITY AND BIJECTIVITY
Selected answers and hints.
1. Counter-examples to surjectivity and injectivity are recorded in the penultimate and final columns respectively.
Domain Codomain f (a) Surj? Inj? Bij? Element a1 6= a2 but
A
B
not hit f (a1 ) = f (a2 )
R
R
sin(a)
no
no
no
2
0, π
R
[−1, 1]
sin(a) yes
no
no
none
0, π
[− π2 , π2 ]
[−1, 1]
sin(a) yes
yes yes
none
none
R
R
a4
no
no
no
−1
−1, 1
4
R≥0
R
a
no
yes no
−1
none
R
R≥0
a4
yes
no
no
none
−1, 1
R≥0
R≥0
a4
yes
yes yes
none
none
2
Q≥0
Q≥0
a
no
yes no
2
none
1
Q\{0}
Q\{0}
yes
yes
yes
none
none
a
C
R
|a|
no
no
no
−1
1, i
C
C
a
yes
yes yes
none
none
The inverses asked for are (c) g : [−1, 1] → [− π2 , π2 ], g(a) = sin−1 (a); (g)
1
g : R≥0 → R≥0 , g(a) = a 4 ; (i) g : Q\{0} → Q\{0}, g(a) = a1 ; (k) g : C → C,
g(a) = a.
2. Let a1 , a2 ∈ Q be such that f (a1 ) = f (a2 ). Then
a31 +1
a31 −2
(a32 − 2)(a31 + 1) = (a31 − 2)(a32 + 1), so a32 a31 + a32 − 2a31 − 2 =
Hence a31 = a32 and therefore a1 = a2 . Thus f is injective.
a32 +1
. Hence
a32 −2
3
3
3
a1 a2 + a1 − 2a32 − 2.
=
We find that f is not surjective, because there is no a ∈ Q with f (a) = 1. (If
f (a) = 1 then a3 + 1 = a3 − 2 which is impossible).
a2 −3
3. Let a1 , a2 ∈ R\{2} be such that f (a1 ) = f (a2 ). Then aa11 −3
−2 = a2 −2 . Hence
(a2 −2)(a1 −3) = (a1 −2)(a2 −3), so a2 a1 −2a1 −3a2 +6 = a1 a2 −2a2 −3a1 +6.
Therefore a1 = a2 . Thus f is injective.
Suppose that there exists a ∈ R\{2} such that f (a) = 1. Then a−3
a−2 = 1.
Hence a − 2 = a − 3, which is impossible, so no such a exists. Hence f is not
surjective, so not bijective.
The function f : R\{2} → R\{1} given by the same rule is bijective. To see
this, the same argument as above shows that f is injective. To show that f
is surjective, let b ∈ R\{1} and let a = 2b−3
b−1 (obtained by setting f (a) = b
and rearranging to make a the subject). Note that a 6= 2, as otherwise
2b − 3 = 2b − 2, which is impossible, so a ∈ R\{2}. Now
2b−3
−3
b−1
(2b − 3) − 3(b − 1)
2b − 3
=
= b.
f (a) = f
=
2b−3
b−1
(2b − 3) − 2(b − 1)
−2
b−1
Thus f is surjective.
4. (a) Suppose that a1 , a2 ∈ (1, ∞) such that f (a1 ) = f (a2 ). Then
a2 +1
.
a22 +1
(a21 a2
a1 a22 )
(a21
a22 )
a1 +1
a21 +1
=
Rearranging, we get
−
+
−
− (a1 − a2 ) = 0.
Each bracket on the left-hand side has a factor of (a1 − a2 ), and we
arrive at (a1 − a2 )(a1 a2 + a1 + a2 − 1) = 0. As a1 , a2 > 1, it follows that
a1 a2 + a1 + a2 − 1 > 0 so we must have a1 = a2 and f is injective.
√
(b) If a1√
, a2 > 2 − 1, then a1 a2 + a1 + a2 − 1 > 0 so, as above,
√ f is injective
on ( 2 − 1, ∞). As there’s a local maximum at x = 2 − 1, it follows
that we lose injectivity if we try to extend the domain further.