Suppose f:X → Y

If A ⊆ X, Does the following hold true?

f(X\A) = f(X) \ f(A).

f: (-4,4) → $\R$

f(x) = x².

A = [0,4] , X\A = [-4,0]

f(X) = [0,16]

f(A) = (0,16]

f(X) \ f(A) = {0}

f(X\A) = [0,16].

f(X) \ f(A) ⫋ (X\A)

Suppose f:X → Y

If A ⊆ X, Does the following hold true?

f(X\A) $\subseteq$ f(X\A).

Let y ∈ f(X) \ f(A).

=> y ∈ f(X) but y ∉ f(A).

y ∈ f(X) => there exists x∈X such that y=f(x).

y ∉ f(A) => there does not exist any a∈A such that y=f(a).

∴ x ∈ X but x ∉ A.

∴ f(x) ∈ f(X\A).

Thus f(X) \ f(A) $\subseteq$ f(X\A).

Let f:X → Y We say that f is 1-1 or injective

if f(x₁) = f(x₂)

=> x₁ =x₂.

(i) f:[-4,4] → $\R$

f(x) = x².

Here, f(1) = f(-1)

Then f is not one-one.

(ii) f:$\R$ → $\R$

f(x) = x³.

Show that f is one - one function.

Suppose

f(x₁) = f(x₂)

=> x₁³ = x₂³

Taking cube roots on both sides

(x₁³)^1/3 = (x₂³)^1/3

=> x₁ = x₂

∴ f is 1-1.

X = {1,2,3,4,5}

Y = {3,4,5,6,7,8}

f:X → Y as f(x) = x+2.

Thus f is 1-1.

A function f:X → Y is said to be Onto or surjective

if codomain (f) = range (f).

f:[-4,4] → $\R$..

f(x) = x².

Co Domain (f) = [0,16].

Thus if y<0 or y>16, then

There does not exists any x ∈ [-4,4] such that f(x) =y.

∴ f is not onto

f:(0,∞) → (0,∞)

f(x) = 1/x.

Show that f is onto function.

Let y ∈ (0,∞).

Claim there exists x ∈ (0,∞) such that

f(x) =y.

Suppose f(x) = y

i.e, 1/x = y.

=> x = 1/y.

Choose x as 1/y.

Then f(x) = 1/x = 1/(1/y)=y

∴ f is onto.

Suppose f:x → y and g:y → z.

The composition of f & g, denoted, gof is defined as follows:

gof:x → z

given by (gof)(x) = g(f(x)).

f:$\R$ → $\R$

f(x) = x²

g:$\R$ → $\R$

g(x) = x³.

Show that fog = gof.

fog(x) = f(g(x))

= f(x³) = (x³)² = x⁶

gof(x) = g(f(x))

= g(x²) = (x²)³ = x⁶.

=> fog = gof

f:$\R$ → $\R$

f(x) = sinx

g:$\R$ → $\R$

g(x) = x²

Show that gof ≠ fog.

fog(x) = f(g(x)) = f(x² = sin(x²).

gof(x) = g(f(x)) = g(sinx) = sin²x.

=> gof ≠ fog.

Let, A = {1,2,3,4,5}

B = {0,1,4,9,10,16,20,25,30}

C = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}

f:A → B

f(a) = a²

g: B → C

$g(b)= \begin{cases}\sqrt{b} & \text { if } b \text { is a perfect } \\ b / 2 & \text { if } b \text { is not } a \text { square. }\end{cases}$

Find gof(a) ?

gof : A → C

gof(a) = g(f(a))

= g(a²)

= √a² = a.

Thus gof(a) = a.

The following statements are equivalent.

i) f(A∩B) = f(A) ∩ f(B)

ii) f is 1-1

iii) for any two disjoint set A&B f(A) & f(B) are disjoint.

ii) => i) Suppose that f is 1-1.

WE know that

f(A∩B) ≤ f(A) ∩ f(B).

Let y∈f(A) ∩ f(B)

=> y∈f(A) and y∈f(B).

y∈f(A) => there exists a∈A such that y=f(a).

y∈f(B) => there exists b∈B such that y=f(b).

∴ f(a) = y = f(b)

Since f is 1-1

=> a=b i.e,a∈ A∩B.

∴ y = f(x) ∈ f(A∩B).

ii) => iii) Suppose that f(A∩B) = f(A) ∩ f(B).

Let A & B be disjoint subsets of x.

i.e, A∩B = φ.

∴ f(A∩B) = f(φ) = φ.

f(A) ∩ f(B) = f(A∩B) = φ.

i.e, f(A) & f(B) are disjoint.

iii) => ii) Suppose that f takes disjoint sets to disjoint sets.

Let x₁, x₂ ∈ X. suppose that

f(x₁) = f(x₂)

Let us assume to the contrary that x₁ ≠ x₂, i.e, f is not 1-1.

x₁ ≠ x₂ => {x₁} & {x₂} are disjoint.

=>f({x₁}) ≠ f({x₂}).

=>{f(x₁)} ≠ {f(x₂)}.

=> f(x₁) ≠ f(x₂). which is a contradiction.

Thus f is 1-1.

Let f: A → B. Then f is 1-1 <=> there exists a function g:B → A such that

i) gof = id_A &

ii) g is onto.

Suppose that f is 1-1.

Define g:B → A

as, $g(b)= \begin{cases}a \text { if } b=f(a) \\ \text { a’ ,otherwise }\end{cases}$

i) gof (a) = g(f(a)) = a = id_A(a)

ii) Let a∈A. Let b= f(a). Then g(b) = g(f(a)) = a.

Thus f is onto.

Suppose ∃ an onto function g:B → A

such that gof = id_A.

Suppose that f(a₁) = f(a₂).

=> g(f(a₁)) = g(f(a₂))

=> (gof)(a₁) = (gof)(a₂)

=> id_A (a₁) = id_A (a₂)

=> a₁ = a₂

Thus f is 1-1.

Let f: A → B.

Then f is onto <=> there exists a function g:B → A such that

i) fog = id_B

ii) g is 1-1.

Suppose f is onto.

for each b∈ B,

A_b = {a∈A : f(a) = b}.

Fix a unique a_b∈ A_b.

Define g: B → A as

g(b) = a_b.

i)fog (b) = f(g(b)) = f(a_b = b = id_B(b).

ii)Suppose g(b₁) = g(b₂)

=>f(g(b₁)) = f(g(b₂))

=>(fog) (b₁) = (fog) (b₂)

=>b₁ = b₂

Thus g is 1-1.

Suppose there exists a 1-1 function g:B → A such that fog = id_b.

Let b ∈ B. Let a = g(b).

f(a) = f(g(b)) = (fog) (b)

= id_b (b) = b.

Thus f is onto.