[PDF] A function is bijective if and only if has an inverse PDF has-inverse-is-bijective.pdf

30 nov 2015 · We say that f is injective if whenever f(a1) = f(a2) for some a1,a2 ∈ A, then a1 = a2 We say that f is bijective if it is both injective and surjective Let f : A → B be bijective Then f has an inverse

A function if surjective (onto) if every element of Recap: Bijectivity ○ A function is bijective if it is both surjective and injective A B Left Inverse of a Function

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The function f : A → B has an inverse if and only if it is a bijection Proof There are two things to prove here Firstly we must show that if f has an inverse then it is

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14 18 Definition (Bijective ) Let A, B be sets A function f : A → B is called bijective if and only if f is both injective and surjective

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f is called onto (surjective) if f (A) = B 3 f is called bijective (textbook notation: one -to-one correspondence) if f is both

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will prove that the function g ∘ f : A → C is also injective B be an invertible function and let f-1 be its inverse We need to prove that f is a bijection, so we will

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Let f : A → B be a function Then the inverse relation f-1 is a function from B to A if and only if f is bijective Furthermore, if f is bijective, then f-1 is also bijective

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A function is invertible if and only if it is bijective Proof Suppose that the function f : A → B is invertible and let f−1 be its inverse First we show that f is injective

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ii) Function f has a left inverse iff f is injective iii) Function f has a inverse iff f is bijective Proof Let A and B be non-empty sets and f : A → B a function i) ⇒

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3 Bijectivity and Inverses Now, let's consider the situation with injectivity and surjectivity pictorially When we say a function is injective, what we mean is that if

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A function is bijective if and only if has an inverse

November 30, 2015

Denition 1.Letf:A!B. We say thatfissurjectiveif for allb2B, there exists ana2A such thatf(a) =b. We say thatfisinjectiveif wheneverf(a1) =f(a2)for somea1;a22A, then a

1=a2. We say thatfisbijectiveif it is both injective and surjective.

Denition 2.Letf:A!B. A functiong:B!Aisthe inverse offiffg= 1Band gf= 1A. Theorem 1.Letf:A!Bbe bijective. Thenfhas an inverse. Proof.Letf:A!Bbe bijective. We will dene a functionf1:B!Aas follows. Letb2B. Sincefis surjective, there existsa2Asuch thatf(a) =b. Letf1(b) =a. Sincefis injective, thisais unique, sof1is well-dened. Now we much check thatf1is the inverse off. First we will show thatf1f= 1A. Let a2A. Letb=f(a). Then, by denition,f1(b) =a. Thenf1f(a) =f1(f(a)) =f1(b) =a. Now we will show thatff1= 1B. Letb2B. Leta=f1(b). Then, by denition,f(a) =b. Thenff1(b) =f(f1(b)) =f(a) =b.Theorem 2.Letf:A!Bhave an inverse. Thenfis bijective.

Proof.Letf:A!Bhave an inversef1:B!A.

First, we will show thatfis surjective. Supposeb2B. Leta=f1(b). Thenf(a) = f(f1(b)) =ff1(b) = 1B(b) =b. Sofis surjective. Now, we will show thatfis injective. Leta1;a22Abe such thatf(a1) =f(a2). We will show a

1=a2. Letb=f(a1). Leta=f1(b). Then

2= 1A(a2)

=f1f(a2) =f1(f(a2)) =f1(b) =a:

But at the same time,

1= 1A(a1)

=f1f(a1) =f1(f(a1)) =f1(f(a2)) =f1(b) =a: Thereforea1=a2and we have shown thatfis injective.1quotesdbs_dbs20.pdfusesText_26

[PDF] [PDF] A function is bijective if and only if has an inverse