<h2 id="successrelations--functions-br-lecture-5success"><Success>Relations & functions </br> lecture-5</Success></h2> <div style='float: left; width:50%;'>

### <Success>Is f(X\A) = f(X) \ f(A) ? </Success> <div style='float: left; width:100%;font-size:30px;'> <p>Suppose f:X → Y</p> <p>If A ⊆ X, Does the following hold true?</p> <p>f(X\A) = f(X) \ f(A).</p> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>f: (-4,4) → $\R$</p> <p>f(x) = x².</p> <p>A = [0,4] , X\A = [-4,0]</p> <p>f(X) = [0,16]</p> <p>f(A) = (0,16]</p> <p>f(X) \ f(A) = {0}</p> <p>f(X\A) = [0,16].</p> <p>f(X) \ f(A) ⫋ (X\A)</p> </div>

### <Success>Is f(X)\f(A) ⊆ f(X\A) ?</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Suppose f:X → Y</p> <p>If A ⊆ X, Does the following hold true?</p> <p>f(X\A) $\subseteq$ f(X\A).</p> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Let y ∈ f(X) \ f(A).</p> <p>=> y ∈ f(X) but y ∉ f(A).</p> <p>y ∈ f(X) => there exists x∈X such that y=f(x).</p> <p>y ∉ f(A) => there does not exist any a∈A such that y=f(a).</p> <p>∴ x ∈ X but x ∉ A.</p> <p>∴ f(x) ∈ f(X\A).</p> <p>Thus f(X) \ f(A) $\subseteq$ f(X\A).</p> </div>

#### <Success>One - one function (injective function)</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Let f:X → Y We say that f is <strong>1-1 or injective</strong></p> <p>if f(x₁) = f(x₂)</p> <p>=> x₁ =x₂.</p> </div>

### <Success>Example</Success> <div style='float: left; width:100%;font-size:30px;'> <p>(i) f:[-4,4] → $\R$</p> <p>f(x) = x².</p> <p>Here, f(1) = f(-1)</p> <p>Then f is not one-one.</p> </div>

### <Success>Example</Success> <div style='float: left; width:100%;font-size:30px;'> <p>(ii) f:$\R$ → $\R$</p> <p>f(x) = x³.</p> <p>Show that f is one - one function.</p> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Suppose</p> <p>f(x₁) = f(x₂)</p> <p>=> x₁³ = x₂³</p> <p>Taking cube roots on both sides</p> <p>(x₁³)^1/3 = (x₂³)^1/3</p> <p>=> x₁ = x₂</p> <p>∴ f is 1-1.</p> </div>

### <Success>Example</Success> <div style='float: left; width:50%;font-size:30px;'> <p>X = {1,2,3,4,5}</p> <p>Y = {3,4,5,6,7,8}</p> <p>f:X → Y as f(x) = x+2.</p> <p>Thus f is 1-1.</p> </div> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/f667c941-dba4-4567-1741-ecdf654e8200/public"></p> </div> <div style='float: right; width:49%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/fbbad76c-bb33-4236-02e3-303f22852200/public"></p> </div>

#### <Success>Onto function (surjective function)</Success> <div style='float: left; width:100%;font-size:30px;'> <p>A function f:X → Y is said to be <strong>Onto or surjective</strong></p> <p>if codomain (f) = range (f).</p> </div>

### <Success>Example</Success> <div style='float: left; width:100%;font-size:30px;'> <p>f:[-4,4] → $\R$..</p> <p>f(x) = x².</p> <p>Co Domain (f) = [0,16].</p> <p>Thus if y<0 or y>16, then</p> <p>There does not exists any x ∈ [-4,4] such that f(x) =y.</p> <p>∴ f is not onto</p> </div>

### <Success>Example</Success> <div style='float: left; width:100%;font-size:30px;'> <p>f:(0,∞) → (0,∞)</p> <p>f(x) = 1/x.</p> <p>Show that f is onto function.</p> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Let y ∈ (0,∞).</p> <p><strong>Claim</strong> there exists x ∈ (0,∞) such that</p> <p>f(x) =y.</p> <p>Suppose f(x) = y</p> <p>i.e, 1/x = y.</p> <p>=> x = 1/y.</p> <p>Choose x as 1/y.</p> <p>Then f(x) = 1/x = 1/(1/y)=y</p> <p>∴ f is onto.</p> </div>

### <Success>Composition function</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Suppose f:x → y and g:y → z.</p> <p>The composition of f & g, denoted, gof is defined as follows:</p> <p>gof:x → z</p> <p>given by (gof)(x) = g(f(x)).</p> </div>

### <Success>Example</Success> <div style='float: left; width:100%;font-size:30px;'> <p>f:$\R$ → $\R$</p> <p>f(x) = x²</p> <p>g:$\R$ → $\R$</p> <p>g(x) = x³.</p> <p>Show that <strong>fog = gof</strong>.</p> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p><strong>fog(x) = f(g(x))</strong></p> <p>= f(x³) = (x³)² = x⁶</p> <p><strong>gof(x) = g(f(x))</strong></p> <p>= g(x²) = (x²)³ = x⁶.</p> <p>=> <strong>fog = gof</strong></p> </div>

### <Success>Example</Success> <div style='float: left; width:100%;font-size:30px;'> <p>f:$\R$ → $\R$</p> <p>f(x) = sinx</p> <p>g:$\R$ → $\R$</p> <p>g(x) = x²</p> <p>Show that gof ≠ fog.</p> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>fog(x) = f(g(x)) = f(x² = sin(x²).</p> <p>gof(x) = g(f(x)) = g(sinx) = sin²x.</p> <p>=> gof ≠ fog.</p> </div>

### <Success>Example</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Let, A = {1,2,3,4,5}</p> <p>B = {0,1,4,9,10,16,20,25,30}</p> <p>C = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}</p> <p>f:A → B</p> <p>f(a) = a²</p> <p>g: B → C</p> <p>$g(b)= \begin{cases}\sqrt{b} & \text { if } b \text { is a perfect } \\ b / 2 & \text { if } b \text { is not } a \text { square. }\end{cases}$</p> <p>Find gof(a) ?</p> </div>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;'> <p>gof : A → C</p> <p>gof(a) = g(f(a))</p> <p>= g(a²)</p> <p>= √a² = a.</p> <p>Thus gof(a) = a.</p> </div>

### <Success>Theorem</Success> <div style='float: left; width:100%;font-size:30px;'> <p>The following statements are equivalent.</p> <p>i) f(A∩B) = f(A) ∩ f(B)</p> <p>ii) f is 1-1</p> <p>iii) for any two disjoint set A&B f(A) & f(B) are disjoint.</p> </div>

### <Success>Proof</Success> <div style='float: left; width:100%;font-size:30px;'> <p>ii) => i) Suppose that f is 1-1.</p> <p>WE know that</p> <p>f(A∩B) ≤ f(A) ∩ f(B).</p> <p>Let y∈f(A) ∩ f(B)</p> <p>=> y∈f(A) and y∈f(B).</p> <p>y∈f(A) => there exists a∈A such that y=f(a).</p> <p>y∈f(B) => there exists b∈B such that y=f(b).</p> </div>

### <Success>Proof</Success> <div style='float: left; width:100%;font-size:30px;'> <p>∴ f(a) = y = f(b)</p> <p>Since f is 1-1</p> <p>=> a=b i.e,a∈ A∩B.</p> <p>∴ y = f(x) ∈ f(A∩B).</p> </div>

### <Success>Proof</Success> <div style='float: left; width:100%;font-size:30px;'> <p>ii) => iii) Suppose that f(A∩B) = f(A) ∩ f(B).</p> <p>Let A & B be disjoint subsets of x.</p> <p>i.e, A∩B = φ.</p> <p>∴ f(A∩B) = f(φ) = φ.</p> <p>f(A) ∩ f(B) = f(A∩B) = φ.</p> <p>i.e, f(A) & f(B) are disjoint.</p> </div>

### <Success>Proof</Success> <div style='float: left; width:100%;font-size:30px;'> <p>iii) => ii) Suppose that f takes disjoint sets to disjoint sets.</p> <p>Let x₁, x₂ ∈ X. suppose that</p> <p>f(x₁) = f(x₂)</p> <p>Let us assume to the contrary that x₁ ≠ x₂, i.e, f is not 1-1.</p> <p>x₁ ≠ x₂ => {x₁} & {x₂} are disjoint.</p> <p>=>f({x₁}) ≠ f({x₂}).</p> <p>=>{f(x₁)} ≠ {f(x₂)}.</p> <p>=> f(x₁) ≠ f(x₂). which is a contradiction.</p> <p>Thus f is 1-1.</p> </div>

### <Success>Theorem</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Let f: A → B. Then f is 1-1 <=> there exists a function g:B → A such that</p> <p>i) gof = id_A &</p> <p>ii) g is onto.</p> </div>

### <Success>Proof</Success> <div style='float: left; width:50%;font-size:30px;'> <p>Suppose that f is 1-1.</p> <p>Define g:B → A</p> <p>as, $ g(b)= \begin{cases}a \text { if } b=f(a) \\ \text { a’ ,otherwise }\end{cases}$</p> <p>i) gof (a) = g(f(a)) = a = id_A(a)</p> <p>ii) Let a∈A. Let b= f(a). Then g(b) = g(f(a)) = a.</p> <p>Thus f is onto.</p> </div> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/09819461-52b0-45b6-2e17-ce578f1f8400/public"></p> </div> <div style='float: right; width:49%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/04c23327-5cd2-4c71-388a-fe664e6b4f00/public"></p> </div>

### <Success>Proof</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Suppose ∃ an onto function g:B → A</p> <p>such that gof = id_A.</p> <p>Suppose that f(a₁) = f(a₂).</p> <p>=> g(f(a₁)) = g(f(a₂))</p> <p>=> (gof)(a₁) = (gof)(a₂)</p> <p>=> id_A (a₁) = id_A (a₂)</p> <p>=> a₁ = a₂</p> <p>Thus f is 1-1.</p> </div>

### <Success>Theorem</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Let f: A → B.</p> <p>Then f is onto <=> there exists a function g:B → A such that</p> <p>i) fog = id_B</p> <p>ii) g is 1-1.</p> </div>

### <Success>Proof</Success> <div style='float: left; width:60%;font-size:30px;'> <p>Suppose f is onto.</p> <p>for each b∈ B,</p> <p>A_b = {a∈A : f(a) = b}.</p> <p>Fix a unique a_b∈ A_b.</p> </div> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/82c0656c-824b-4081-5a15-67b9dccb9c00/public"></p> </div> <div style='float: right; width:39%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/0ec7c06a-229e-4612-7ed8-046b1f10b700/public"></p> </div>

### <Success>Proof</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Define g: B → A as</p> <p>g(b) = a_b.</p> <p>i)fog (b) = f(g(b)) = f(a_b = b = id_B(b).</p> <p>ii)Suppose g(b₁) = g(b₂)</p> <p>=>f(g(b₁)) = f(g(b₂))</p> <p>=>(fog) (b₁) = (fog) (b₂)</p> <p>=>b₁ = b₂</p> <p>Thus g is 1-1.</p> </div>

### <Success>Proof</Success> <div style='float: left; width:100%;font-size:30px;'> <p>Suppose there exists a 1-1 function g:B → A such that fog = id_b.</p> <p>Let b ∈ B. Let a = g(b).</p> <p>f(a) = f(g(b)) = (fog) (b)</p> <p>= id_b (b) = b.</p> <p>Thus f is onto.</p> </div>

### <Success>Thank you</Success> <div> </div>