1. Today’s topics:
Proof that a given function is injective or surjective.
Composite functions
Properties of the composite functions.
Inverse functions.
Proofs with functions.

2. Let A =R{1} and f : A A be defined by the formula:
Prove that f is injective and surjective.
1) injective: Let x and y be arbitrary elements of A such that
f (x) = f (y ), (1). We must show that (1) implies x = y
by assumption
since (x 1)0 and (y 1)0
by algebra

3. 2) surjective: for any y there exists x A , such that
f (x)=y .
Take arbitrary y A and find x A from f (x)=y

4. Composition of functions a A f B
f (a)=b g C
g (b)=c
gf
(Note, that in the context of function composition, the order
in which the functions appear is backward, i. e. the rightmost
function is applied first)
Theorem. Let f : AB and g: BC be two functions.
The composition of f and g as relations defines a function
gf : A C, such that gf (a)= g(f(a)).

5. Proof. We need to prove that composite relation gf is a function.
For this we must prove two things:
1) the composition gf relates each element aA to some c C.
2) an element c C assigned to a by gf is unique, so we can
denote it c= g(f (a)).
1) existence: Let b=f (a) B. Let c=g(b) C. So, by the definition
of composition of relations, (a, c)gf . Thus, cC [(a, c)gf ]  a A g C
g (b)=c f B
f (a)=b
gf

6. 2) uniqueness: Suppose (a, c1)gf and (a, c2)gf .
Then by the definition of composition, b1B, such that
(a, b1)f and (b1, c1)g,
and b2B, such that (a, b2)f and (b2, c2)g.
Since f is a function, there may be only one b B, such that
(a, b)f, so b1= b2.
Since g is a function, (b, c1)g and (b, c2)g imply that c1= c2. a c1 c2 b1 f g b2
Thus, (gf )(a)= c= g (b) = g (f (a)).

7. Properties of the composite function
How the injective (surjective) property of gf depends
on properties of f and g f g g f • g f • f g

8. Theorem. Consider functions f : AB and g: B C, and
the composition gf : AC.
a) If both f and g are injective, then gf is injective.
b) If both f and g are surjective, then gf is surjective.
c) If both f and g are bijective, then gf is bijective.
Proof. a) Let a1 and a2 be arbitrary elements of A such that
(gf )(a1) = (gf )(a2). By the previous Theorem, it implies that
g (f (a1)) = g (f (a2)). Since g is injective it implies that
f (a1) = f (a2), and since f is injective by assumption,
a1 = a2. a1 a2
g(f (a1))=g(f (a2)). f g
f (a1)
f (a2) A B C

9. b) Take an arbitrary element cC.
Since g is surjective, there exists bB such that g(b)=c.
Similarly, since f is surjective,there exists aA such that f (a)=b.
Then (gf )(a) = g(f(a)) = g(b) = c.
So, gf is surjective. C c  g B b f A a
gf
g surjective: c, b (g(b)=c)
f surjective: b, a (f (a)=b)

10. c) If both f and g are bijective, then both f and g are injective and
surjective. By parts a) and b) gf is injective and surjective,
so gf is bijective.
So, injective (surjective) properties of two functions, f and g
are sufficient condition for injective (surjective) property of the
composite function. The question arises whether both conditions
are necessary as well.
Let f : AB and g: B C be two functions and
the composition gf : AC is injective.
What properties of of f and g can be implied?
What properties of f and g can be implied if gf is surjective?

11. Example. Let f : AB and g: B  C be two functions such that
the composite function gf : AC is injective.
Prove or disprove that both f and g are injective.
Injective property of gf : AC implies
if (gf )(x)=(gf )(y), then x = y.
Let’s prove by contradiction that f must be injective.
For this assume that gf is injective, but f is not injective.
Then we can find x, y A such that x y and f (x)=f (y).
By taking composite function for these x and y we have that
g(f (x)) = g(f (y)) while x y.
But this results to contradiction with injective property of composite function.
(gf )(x) x y f g
x y
g( f (x))= g( f (y))
(gf )(y)
gf injective  f injective

12. What about gf injective  g injective ?
This implication can be disproved by the following
counterexample: f g
g is not injective, but gf is!
If f was surjective, then we would not be able to disprove
that gf injective  g injective!

13. Example: f : [0, ]  [0, ], f (x)=x/2.
g : [0, ]  [0, 1], g (y)=sin(y), not injective
g(y) = sin(y): [0, ]  [0, 1]
gf =sin(x/2): [0, ]  [0, 1]
gf =sin(x/2): [0, ]  [0, 1], is injective, but g is not.

14. Example. Let f : AB and g: B  C be two functions such that
the composite function gf : AC is surjective.
Prove or disprove that both f and g are surjective.
Let’s prove that if gf is surjective then g is surjective.
Surjective property of gf implies that for any zC there exists
x A such that (gf )(x)=z. It means that g(f (x))=z.
Since f is a function, there exists a unique element y B such that
y=f (x). But g(f (x))= g(y)= z. z x
gf 
y=f (x)
Thus, for any zC there exists y B such that g(y)= z.
Than means that g is surjective.

15. / The following example shows that gf is surjective  f is surjective
If we knew that g were injective, we could prove that
gf is surjective  f is surjective !

16. Example. Let f : RR, and g : RR denote two functions,
where R is the set of real numbers. Suppose gf (x)=2x+1 for
all x R ( i. e. the composite function is both surjective and
injective).
a) Prove that the function f is injection.
Let f (x)= f (y), we need to show that x = y.
Since g is a function on a set of real numbers, f (x)= f (y) implies
g(f (x))= g (f (y)), i. e. 2x+1 = 2y+1 (by assumption gf (x)=2x+1).
So, it yields x = y . QED.
b) Prove the function g is a surjection.
Take arbitrary z R, we need to show that y R such that g(y)=z.
But for any z R there always exist x, y R, x = (z1)/2 and y=f (x),
so that g(y) = g (f (x)) = 2(z1)/2+1=z

17. Inverse function. Suppose the relation R  A B is a function.
The inverse relation was defined as:
R-1 = { (b, a) | (a, b) R }  BA
Is R-1 a function?
R-1 R
R is a function
R-1 is not a function. Why?

18. Theorem. Let f : A  B be a bijection. Then the inverse relation  R
R is a function 
R-1
R-1 is not a function. Why?
Theorem. Let f : A  B be a bijection. Then the inverse relation
of f , defined from B to A as {(b, a)| bB, aA and f (a)=b}, is
a function from B to A such that gf (a)=a for all a A and
fg (b)=b for all b B.
The function g is called the inverse function of f and is denoted f –1.

19. Proof. Since f : A  B is a bijection, for each bB there exists
one and only one element aA such that f (a) =b .
Thus, the relation from B to A, relating bB to its pre-image aA
under f , is a function.
That is we defined a function
g : BA such that for any bB, g(b) =a iff f (a)=b.
Thus, gf (a)= g(f (a)) = a
Similarly, fg (b)=f (g (b))=b.