FUNCTION M05 THEOREMS WITH PROOFS FOR BOARD LEVEL

1. FUNCTIONS
FUNCTIONS

2. FUNCTIONS
THEOREMS

3. FUNCTIONS
Theorem(1): If f:AB, g:BC & h:CD are three functions.
Then prove that ho(gof)=(hog)of i.e., composition of
functions is associative.
Proof
Part-1
Given that f:AB and g:BC
Now, gof:AC and h:CD
Also, g:BC and h:CD
 gof:AC
ho(gof) : A  D
 hog:BD
Do you
know gof?
Do you know
ho(gof)?
Do you know
hog?

4. FUNCTIONS
Thus, ho(gof) and (hog)of have the same domain A.
Now f:AB and hog:BD
 (hog)of:AD
Part-2 Let aA
[ho(gof)] (a) = h[(gof) (a)]
= h[g(f (a))]
= (hog)[f (a)]
= (hog)of (a)
Hence, from part-1 and part-2 we conclude that
ho(gof)=(hog)of aA
Do you know
(hog)of?
ho(gof) : A  D

5. FUNCTIONS
Theorem(2): If f:AB, g:BC are two bijective functions
then prove that gof:AC is also a bijective function
Proof:
Given that f, g are bijections.
(i) To prove that gof:AC is one-one
Let a1,a2A, then f(a1),f(a2)B
Hence, g(f(a1)), g(f(a2))C
Let (gof)(a1) = (gof)(a2)
g(f(a1)) = g(f(a2))
Hence f, g are both one-one
and onto functions
Bijection means it is
both one-one & on-to
To prove gof:AC is
bijective, firstly we
prove it is one-one
Since f:AB
is a function
Since g:BC
is a function
Since f and g are functions we know that gof:AC is a
function
These are
images of a1, a2

6. FUNCTIONS
(ii) To prove that gof:AC is onto
Let c be any element in co-domain C
Since g:BC is onto
Now, bB and f:AB is onto
bB  aA such that f(a)=b(2)
Thus gof:AC is one-one.
f(a1) = f(a2) (∵ g is one-one)
a1 = a2 (∵ f is one-one)
Later we prove
gof:AC onto
function
cC  bB such that g(b)=c(1)
By the
definition of
onto function

7. FUNCTIONS
Hence, gof:AC is both one one and onto (i.e.,) gof is
bijective function.
As c is arbitrary, every element in C has pre-image in
domain A under gof. Thus gof : A C is onto.
Hence proved.
From eq(1)
for cC, (gof)(a)=c for some aA
c = g(b)
= g(f(a))
c = (gof)(a)
From eq(2)
b=f(a)

8. FUNCTIONS
Theorem(3): If f : A  B, g : B  C are two bijective
functions then prove that (gof)-1 = f-1og-1
Proof
Given that f : A  B, g : B  C are bijective
Part 1
i.e., A
𝐟
B
𝐠
C are bijections
gof:
(gof)-1:CA is inverse of (gof)
Also f-1:BA, g-1:CB are both bijections.
Firstly we prove
(gof)-1, f-1og-1 have
the same domain
AC is a bijection
If f,g are bijections,
f-1, g-1 are also
bijections.

9. FUNCTIONS
Thus (gof)-1 and f-1 og-1 both have the same domain “C”.
i.e., C
𝐠−𝟏
B
𝐟−𝟏
𝐀 are bijections.
(f-1og-1):CA
Part 2
Let c be any element in C. By data g:BC is a bijection
(i.e.,) g:BC is onto.
Hence, for cC  a unique bB such that g(b) = c
b=g-1(c)
and (gof)-1:CA
By observing
(gof)-1, f-1og-1, they
have the same domain
C
By the
definition of
onto function

10. FUNCTIONS
Also, f:AB is a bijection (i.e.,) f:AB is onto.
Hence, bB  a unique aA such that f(a) = b
Now c = g(b)
a = f-1(b)
= g(f(a))
c= (gof)(a)
 (gof)-1(c) = a(1)
Also, (f-1og-1) (c)= f-1(g-1(c))
= f-1(b)
= a
(f-1og-1) (c) = a  (2)
 From (1) & (2)
(gof)-1 (c) = (f-1og-1) (c),  cC
We know
g(b) = c
We know
g-1(c)=b

11. FUNCTIONS
Theorem(4): If f:AB is a function and IA & IB are identity
functions on A,B respectively then prove that foIA=f=IBof
Proof
(i) To prove that foIA=f
Part 1 We know that IA:AA.
 foIA :
IA(a) = a  aA
Given that
f: A  B. Thus A
𝐈𝐀
A
𝐟
𝐁
Thus foIA and f both have the same domain A
Firstly we prove
foIA, f have the
same domain
A  B
By observing
foIA, f they have
the same domain
A

12. FUNCTIONS
Part 2
Hence, from Part-1 and Part-2 we conclude that foIA=f
Let aA, then (foIA) (a) = f(IA(a))
(ii) To prove that IBof = f
Since f:AB and IB:BB
 IB(b) = b,  bB
(i.e.,) A
𝐟
A
𝐈𝐁
𝐁
 IBof :
= f(a)
Thus f and IBof both have the same domain A
Part 1
We know
[ ∵ IA(a) = a  a A ]
A  B
By observing
f and IBof they
have the same
domain A

13. FUNCTIONS
Hence from part-1 and part-2 we conclude that IBof = f
Hence proved.
Part2
[ ∵ IB(b)=b,  bB and f(a)B]
For any aA, (IBof)(a)=IB(f(a))=f(a)

14. FUNCTIONS
Theorem(5): If f:AB is a bijective function. IA& IB are
identity functions then prove
Proof:
Part 1
Given f:AB is a bijection then we know, f-1:B A is also
a bijection
Now, f-1:BA and f:AB
(i) To prove that fof-1=IB
i) fof-1=IB ii) f-1of=IA
(i.e.,) B
𝐟−𝟏
𝐀
𝐟
𝐁
Firstly, we
prove fof-1, IB
have the same
domain
fof-1: BB

15. FUNCTIONS
Part 2 Let bB. Since f:AB is a bijection  a unique a  A
such that f(a)=bf-1(b)=a
Thus fof-1 and IB both have the same domain B
Also by the definition of identity function, IB(b)=b  b  B
Now, for any bB,
Also by the definition of identity function, IB:BB
(fof-1)(b) =f(f-1(b)) =f(a)=b=IB(b)
Hence, from part-1 and part-2, we conclude that fof-1=IB

16. FUNCTIONS
(ii) To prove that f-1of=IA
Part 1
Since f:AB and f-1:BA
(i.e.,) A
𝐟
𝐁
𝐟−𝟏
𝐀 f-1of: AA
Also by the definition of identity function IA:AA
Thus, f-1of and IA both have the same domain of A

17. FUNCTIONS
Part 2
Also by the definition of identity function
Let a1A since f:AB is a bijection,  a unique b1B
such that f(a1)=b1
IA(a1)=a1  a1  A
Now for any a1A
 f-1(b1) =a1
(f-1of)(a1) =f-1(f(a1)) =f-1(b1) =a1 =IA(a1)
Hence, from part-1 and part-2 we conclude that f-1of = IA

18. FUNCTIONS
As f: A  B is a function so f(a1), f(a2)B
Theorem(6): If f: A  B and g: B  A are two functions
such that gof = IA and fog = IB, then g = f-1
Proof:
First of all, we prove that f: A  B is a bijection.
To show that f is one-one
Let a1,a2A.
Bijection means
it is both
one-one & onto
Firstly, we
prove f is
one-one
Assume f(a1) = f(a2)
g(f(a1)) = g(f(a2))
(∵g:BA is a function)
(gof)(a1) = (gof)(a2)

19. FUNCTIONS
IA(a1) = IA(a2) ( ∵ gof = IA)
a1 = a2
 f: A  B is one-one function.
To show that f is onto:
Let b be any element in B, then as g : BA is a
function,  aA such that g(b) = a
Now b = IB(b)
= (fog)(b)
= f(g(b))
= f(a)
i.e., f(a) = b
We know
fog = IB

20. FUNCTIONS
Thus for any bB, then there exists pre image aA under f.
Hence f is onto
∴ f: A  B is bijection.
f: A  B is bijection, so f-1 exists.
To show that g = f-1
g: B  A is a function.
f-1: B  A is also bijection.
g, f-1 have same domain B.
Let b be any element in B, then as g : BA is a function, 
a  A such that g(b)=a for some a  A

21. FUNCTIONS
Thus, we get g=f-1
f : AB is bijection, bB  aA such that f(a) = b
 g(b)=f-1(b)
a = f-1(b)
Hence proved
We know
g(b)=a

22. FUNCTIONS
Theorem(7): If f:AB, g:BA are two functions and
(i) gof:AC is one-one, then f:AB is to one-one
(ii) gof:AC is on-to, then g:BC is to on-to
Proof i) Let a1,a2 A
Now f(a1)=f(a2)
gof(a1)=gof(a2)
(∵g:AB is a function)
a1=a2 (∵gof is one-one)
Thus, f(a1)=f(a2)
g(f(a1))=g(f(a2))
a1=a2
i.e,f is one-one function.
Hence proved
If we take
f(a1) = f(a2), we get
a1 = a2

23. FUNCTIONS
Thus g:BC is onto as every element in co-domain C has
pre-image in domain B under g.
Hence proved
(ii) Let c be any element in C. As gof:AC is onto, There
exists a A such that
Thus pre-image of c is f(a)B under g
(gof)(a)=c g(f(a))=c

24. FUNCTIONS
Thank you…