1. Proofs involving functions.
Let f: AB be a function. Then for any subset of A, CA, we can
define the set of images for elements of C that we denote as f (C):
f (C) = {f (x) | xC }. A B
CA
f (C) f

2. For any subset of B, DB, we can define the set of pre-images
of elements from D, which we denote as f -1(D):
f -1(D) = {x | xA and f (x)D }. A B
f -1(D)
DB f
Pay attention that f -1(D) is just a notation for the set of pre-images,
which does not assumes the existence of inverse function for f.

3. Example: A = {a, b, c}, B ={x, y, z}, and f (a)=x, f (b)=y, f (c)=y. A B a b c x y z
In this example, function f is neither injective nor surjective.
So, inverse function f -1 does not exist (in other words, inverse
relation of f is not a function). But we can find a set of pre-images
for any subset of B. For example:
f -1({z})=; f -1({x, z})={a}; f -1({x, y}) = {a, b, c}.

4. Questions
Is it always true, that f (A) = B? A B •
f : AB
f (A)
Ans. It is true only if f is surjective. Otherwise f (A)B.

5. Is it always true that f -1(B)=A? A
f : AB B
f -1(B)
Ans. Yes, since f is a function, any element of A has an image
that belongs B. As a result any element of A appears as a pre-image
of some element of B.

6. Is it true that if f(C)=f(D), where C, DA, then C=D? A
f : AB B C
f(C)
=f(D) D
Ans. This is true only if f is injective.

7. Is it true that if f -1(C)=f -1(D), where C, D  B, then C=D? A
f : AB B D C
f -1(C)
=f -1(D)
Ans. This is true only if f is surjective

8. Example. Prove that if f is injective, then XY= implies
f(X)f(Y)=. f B A X
f (X) Y
f (Y)

9. Proof by contradiction. Assume that XY= and
f(X)f(Y), (1).
(1) implies that there exists some common element,
zf(X) and z f(Y), (2).
From the definition of the set f(X) and f (Y) (2) implies that
there exists xX, such that f(x)=z (3)
and there exists yY, such that f(y)=z (4).
So, we have f(x)= f(y)=z, (5)
and since f is injective (5) implies that
x=y, (6).
Since xX and yY , (6) implies that XY in contradiction
with assumption.
The contradiction proves the initial statement.

10. Let f : A  B be a function.
Prove or disprove that for any D  B, f (f 1(D )) =D  A
f : AB B D
f 1(D)
= f (f 1(D ))

11.  A
f : AB B
f 1(D) D
f (f 1(D )) D

12. Suppose R A  A is an equivalence relation. Prove or disprove
that if R is a function, then this function is IA
(identity function on A)
Proof by contradiction. Assume that R is an equivalence relation
on A that defines a function f : A A , and f IA .
It means that there exist a b , such that f (a)=b.
At the same time (a, b) R .
Since R is an equivalence relation, it is reflexive, symmetric
and transitive. It implies, that (b, a) R (symmetric).
By the transitive property (a, b) R and (b, a) R imply (a, a) R
But (a, b) R and (a, a) R means that R is not a function!

13. #2, p. 277
a) If f : A B and (a, b), (a, c )  f , then b = c
b) If f : A B is a one-to one correspondence (injective)
and A and B are finite, then |A|=|B|
c) If f : A B is a one-to one (injective), then f is invertible
( f 1 is a function).
d) If f : A B is invertible ( f 1 is a function), then f is
one-to one (injective).
e) If f : A B is a one-to one (injective) and g, h : B C
with g  f = h  f , then g = h .

14. f) If f : A B and A1, A2  A, then f (A1 A2 ) = f (A1)  f (A2 ) A B
f : A B A1
f (A1 A2 ) A2

15. g) If f : A B and B1, B2  B, then
f 1(B1 B2 ) = f 1(B1)  f 1(B2 )
#6, p Let A = {1, 2, 3, 4, 5} and B = {1, 2, 3, 4, 5, 6}.
How many injective functions f : A B satisfy
a) f (1) = 3 ?
b) f (1) = 3, f (2) = 6 ?

16. Let R  AA be a transitive relation.
Prove or disprove that s(R) is transitive.
Let R  AA be an equivalence relation.
i)Prove that RR is equivalence relation.
ii)Prove or disprove that R and RR induce the same partition of A.
Let f AB and g AB are two relations.
Suppose the relations f, g and f g are functions from A to B
(That is for any element aA there exists a unique element bB,
(a, b)  f . The same for g and f g.) Prove that f = g.
    Let A ={1, 2, 3, 4, 5} {1, 2, 3, 4, 5}, and define R on A by
(x1, y1)R(x2, y2) if x1+ y1= x2+y2.
i) verify that R is equivalence relation.
ii) Determine the equivalence classes [(1, 3)], [(2, 4)], and [(1,1)].
iii) Determine the partition of A induced by R.