Functions - Part 2 Supartha Podder One-to-One and Onto Functions A - - PowerPoint PPT Presentation

Functions - Part 2

Supartha Podder

One-to-One and Onto Functions

A function f : A → B is said to be one-to-one, or an injunction, if and only if f (a) = f (b) implies that a = b for all a, b ∈ A. A function is said to be injective if it is one-to-one. ∀a ∈ A, ∀b ∈ A (f (a) = f (b) → a = b)

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Onto Functions

Onto

A function f from A to B is called onto, or a surjection, if and only if for every element b ∈ B there is an element a ∈ A with f (a) = b. A function f is called surjective if it is onto. ∀y ∈ B ∃x ∈ A (f (x) = y)

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Example

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Example of Bijective Function

Let f : (R+ × R+) → (R+ × R+) be the function defined by f (r, s) = (2r, rs). Proof that f is bijective.

Proof of Injectivity

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Example of Bijective Function

Let f : (R+ × R+) → (R+ × R+) be the function defined by f (r, s) = (2r, rs). Proof that f is bijective.

Proof of Injectivity

Let (a, b), (c, d) ∈ R+ × R+ be arbitrary elements of f ’s domain.

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Example of Bijective Function

Let f : (R+ × R+) → (R+ × R+) be the function defined by f (r, s) = (2r, rs). Proof that f is bijective.

Proof of Injectivity

Let (a, b), (c, d) ∈ R+ × R+ be arbitrary elements of f ’s domain. Assume f (a, b) = f (c, d) (Goal is to prove (a, b) = (c, d).)

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Example of Bijective Function

Let f : (R+ × R+) → (R+ × R+) be the function defined by f (r, s) = (2r, rs). Proof that f is bijective.

Proof of Injectivity

Let (a, b), (c, d) ∈ R+ × R+ be arbitrary elements of f ’s domain. Assume f (a, b) = f (c, d) (Goal is to prove (a, b) = (c, d).) Then, (2a, ab) = (2c, cd) ⇒ 2a = 2c and ab = cd.

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Example of Bijective Function

Let f : (R+ × R+) → (R+ × R+) be the function defined by f (r, s) = (2r, rs). Proof that f is bijective.

Proof of Injectivity

Let (a, b), (c, d) ∈ R+ × R+ be arbitrary elements of f ’s domain. Assume f (a, b) = f (c, d) (Goal is to prove (a, b) = (c, d).) Then, (2a, ab) = (2c, cd) ⇒ 2a = 2c and ab = cd. For two ordered pairs to be equal, both their 1st and 2nd coordinates must be equal.

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Example of Bijective Function

Let f : (R+ × R+) → (R+ × R+) be the function defined by f (r, s) = (2r, rs). Proof that f is bijective.

Proof of Injectivity

Let (a, b), (c, d) ∈ R+ × R+ be arbitrary elements of f ’s domain. Assume f (a, b) = f (c, d) (Goal is to prove (a, b) = (c, d).) Then, (2a, ab) = (2c, cd) ⇒ 2a = 2c and ab = cd. For two ordered pairs to be equal, both their 1st and 2nd coordinates must be equal. Thus, a = c and ab − cd = 0. ⇒ ab − ad = 0 since a = c ⇒ a(b − d) = 0.

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Example of Bijective Function

Let f : (R+ × R+) → (R+ × R+) be the function defined by f (r, s) = (2r, rs). Proof that f is bijective.

Proof of Injectivity

Let (a, b), (c, d) ∈ R+ × R+ be arbitrary elements of f ’s domain. Assume f (a, b) = f (c, d) (Goal is to prove (a, b) = (c, d).) Then, (2a, ab) = (2c, cd) ⇒ 2a = 2c and ab = cd. For two ordered pairs to be equal, both their 1st and 2nd coordinates must be equal. Thus, a = c and ab − cd = 0. ⇒ ab − ad = 0 since a = c ⇒ a(b − d) = 0. So, a = 0 or b = d. Now a = 0 is not possible because of the domain. So b = d. Hence f is injective.

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Example of Bijective Function

Let f : (R+ × R+) → (R+ × R+) be the function defined by f (r, s) = (2r, rs). Proof that f is bijective.

Proof of Surjectivity

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Example of Bijective Function

Let f : (R+ × R+) → (R+ × R+) be the function defined by f (r, s) = (2r, rs). Proof that f is bijective.

Proof of Surjectivity

Let (u, v) ∈ R+ × R+ be an arbitrary element of f ’s co-domain. We want to find at least one element (r, s) ∈ R+ × R+ (in f ’s domain) such that f (r, s) = (u, v).

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Example of Bijective Function

Let f : (R+ × R+) → (R+ × R+) be the function defined by f (r, s) = (2r, rs). Proof that f is bijective.

Proof of Surjectivity

Let (u, v) ∈ R+ × R+ be an arbitrary element of f ’s co-domain. We want to find at least one element (r, s) ∈ R+ × R+ (in f ’s domain) such that f (r, s) = (u, v). Well, f (r, s) = (u, v) ⇔ (2r, rs) = (u, v)

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Example of Bijective Function

Let f : (R+ × R+) → (R+ × R+) be the function defined by f (r, s) = (2r, rs). Proof that f is bijective.

Proof of Surjectivity

Let (u, v) ∈ R+ × R+ be an arbitrary element of f ’s co-domain. We want to find at least one element (r, s) ∈ R+ × R+ (in f ’s domain) such that f (r, s) = (u, v). Well, f (r, s) = (u, v) ⇔ (2r, rs) = (u, v) ⇔ 2r = u and rs = v.

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Example of Bijective Function

Let f : (R+ × R+) → (R+ × R+) be the function defined by f (r, s) = (2r, rs). Proof that f is bijective.

Proof of Surjectivity

Let (u, v) ∈ R+ × R+ be an arbitrary element of f ’s co-domain. We want to find at least one element (r, s) ∈ R+ × R+ (in f ’s domain) such that f (r, s) = (u, v). Well, f (r, s) = (u, v) ⇔ (2r, rs) = (u, v) ⇔ 2r = u and rs = v. ⇔ r = u

2 and s = v r .

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Example of Bijective Function

Let f : (R+ × R+) → (R+ × R+) be the function defined by f (r, s) = (2r, rs). Proof that f is bijective.

Proof of Surjectivity

Let (u, v) ∈ R+ × R+ be an arbitrary element of f ’s co-domain. We want to find at least one element (r, s) ∈ R+ × R+ (in f ’s domain) such that f (r, s) = (u, v). Well, f (r, s) = (u, v) ⇔ (2r, rs) = (u, v) ⇔ 2r = u and rs = v. ⇔ r = u

2 and s = v r .

Thus, s =

v ( u

2 ) = 2v

u .

Now we need to make sure that (r, s) = ( u

2, 2v u ) ∈ R+ × R+.

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Example of Bijective Function

Let f : (R+ × R+) → (R+ × R+) be the function defined by f (r, s) = (2r, rs). Proof that f is bijective.

Proof of Surjectivity

Let (u, v) ∈ R+ × R+ be an arbitrary element of f ’s co-domain. We want to find at least one element (r, s) ∈ R+ × R+ (in f ’s domain) such that f (r, s) = (u, v). Well, f (r, s) = (u, v) ⇔ (2r, rs) = (u, v) ⇔ 2r = u and rs = v. ⇔ r = u

2 and s = v r .

Thus, s =

v ( u

2 ) = 2v

u .

Now we need to make sure that (r, s) = ( u

2, 2v u ) ∈ R+ × R+.

Since u ∈ R+, so is u

2 and since u, v ∈ R+ so is 2u v .

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Example of Bijective Function

Let f : (R+ × R+) → (R+ × R+) be the function defined by f (r, s) = (2r, rs). Proof that f is bijective.

Proof of Surjectivity

Let (u, v) ∈ R+ × R+ be an arbitrary element of f ’s co-domain. We want to find at least one element (r, s) ∈ R+ × R+ (in f ’s domain) such that f (r, s) = (u, v). Well, f (r, s) = (u, v) ⇔ (2r, rs) = (u, v) ⇔ 2r = u and rs = v. ⇔ r = u

2 and s = v r .

Thus, s =

v ( u

2 ) = 2v

u .

Now we need to make sure that (r, s) = ( u

2, 2v u ) ∈ R+ × R+.

Since u ∈ R+, so is u

2 and since u, v ∈ R+ so is 2u v .

Thus f ( u

2, 2v u ) = (r, s) and f is surjective.

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Inverse Functions

Let f be a one-to-one correspondence from the set A to the set B. The inverse function of f , f −1 is the function that assigns to an element b belonging to B the unique element a in A such that f (a) = b. The inverse function of f is denoted by f −1. Hence, f −1(b) = a whenf (a) = b.. Bijectivity = Injectivity + Surjectivity.

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Example of Inverse

Is f : Z2 → Z2 defined by f (x, y) = (x − y, −x) invertible?

Injective:

Let (p, q), (r, s) ∈ Z × Z be arbitrary elements of f ’s domain.

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Example of Inverse

Is f : Z2 → Z2 defined by f (x, y) = (x − y, −x) invertible?

Injective:

Let (p, q), (r, s) ∈ Z × Z be arbitrary elements of f ’s domain. Assume f (p, q) = f (r, s) (Goal is to prove (p, q) = (r, s).)

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Example of Inverse

Is f : Z2 → Z2 defined by f (x, y) = (x − y, −x) invertible?

Injective:

Let (p, q), (r, s) ∈ Z × Z be arbitrary elements of f ’s domain. Assume f (p, q) = f (r, s) (Goal is to prove (p, q) = (r, s).) Then, (p − q, −p) = (r − s, −r)

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Example of Inverse

Is f : Z2 → Z2 defined by f (x, y) = (x − y, −x) invertible?

Injective:

Let (p, q), (r, s) ∈ Z × Z be arbitrary elements of f ’s domain. Assume f (p, q) = f (r, s) (Goal is to prove (p, q) = (r, s).) Then, (p − q, −p) = (r − s, −r) ⇒ p = r and q = s

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Example of Inverse

Is f : Z2 → Z2 defined by f (x, y) = (x − y, −x) invertible?

Injective:

Let (p, q), (r, s) ∈ Z × Z be arbitrary elements of f ’s domain. Assume f (p, q) = f (r, s) (Goal is to prove (p, q) = (r, s).) Then, (p − q, −p) = (r − s, −r) ⇒ p = r and q = s thus, (p, q) = (r, s).

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Example of Inverse

Is f : Z2 → Z2 defined by f (x, y) = (x − y, −x) invertible?

Injective:

Let (p, q), (r, s) ∈ Z × Z be arbitrary elements of f ’s domain. Assume f (p, q) = f (r, s) (Goal is to prove (p, q) = (r, s).) Then, (p − q, −p) = (r − s, −r) ⇒ p = r and q = s thus, (p, q) = (r, s).

Surjectivity:

Let (u, v) ∈ Z × Z be an arbitrary element of f ’s co-domain. We want to find at least one element (r, s) ∈ Z × Z (in f ’s domain) such that f (r, s) = (u, v).

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Example of Inverse

Is f : Z2 → Z2 defined by f (x, y) = (x − y, −x) invertible?

Injective:

Let (p, q), (r, s) ∈ Z × Z be arbitrary elements of f ’s domain. Assume f (p, q) = f (r, s) (Goal is to prove (p, q) = (r, s).) Then, (p − q, −p) = (r − s, −r) ⇒ p = r and q = s thus, (p, q) = (r, s).

Surjectivity:

Let (u, v) ∈ Z × Z be an arbitrary element of f ’s co-domain. We want to find at least one element (r, s) ∈ Z × Z (in f ’s domain) such that f (r, s) = (u, v). By definition of f , (u, v) = (r − s, −r), thus v = −r and u = r − s.

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Example of Inverse

Is f : Z2 → Z2 defined by f (x, y) = (x − y, −x) invertible?

Injective:

Let (p, q), (r, s) ∈ Z × Z be arbitrary elements of f ’s domain. Assume f (p, q) = f (r, s) (Goal is to prove (p, q) = (r, s).) Then, (p − q, −p) = (r − s, −r) ⇒ p = r and q = s thus, (p, q) = (r, s).

Surjectivity:

Let (u, v) ∈ Z × Z be an arbitrary element of f ’s co-domain. We want to find at least one element (r, s) ∈ Z × Z (in f ’s domain) such that f (r, s) = (u, v). By definition of f , (u, v) = (r − s, −r), thus v = −r and u = r − s. ⇒ r = −v and s = −(u + v). And since (u, v) ∈ Z × Z, (r, s) ∈ Z × Z,

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Example of Inverse

Is f : Z2 → Z2 defined by f (x, y) = (x − y, −x) invertible?

Injective:

Let (p, q), (r, s) ∈ Z × Z be arbitrary elements of f ’s domain. Assume f (p, q) = f (r, s) (Goal is to prove (p, q) = (r, s).) Then, (p − q, −p) = (r − s, −r) ⇒ p = r and q = s thus, (p, q) = (r, s).

Surjectivity:

Let (u, v) ∈ Z × Z be an arbitrary element of f ’s co-domain. We want to find at least one element (r, s) ∈ Z × Z (in f ’s domain) such that f (r, s) = (u, v). By definition of f , (u, v) = (r − s, −r), thus v = −r and u = r − s. ⇒ r = −v and s = −(u + v). And since (u, v) ∈ Z × Z, (r, s) ∈ Z × Z, and f (r, s) = (u, v).

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Example of Inverse

Is f : Z2 → Z2 defined by f (x, y) = (x − y, −x) invertible?

Inverse

Now that we have proven bijectivity. Let us find f −1.

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Example of Inverse

Is f : Z2 → Z2 defined by f (x, y) = (x − y, −x) invertible?

Inverse

Now that we have proven bijectivity. Let us find f −1. We know that f (x, y) = (r, s) hence we need to design the inverse, f −1(r, s) = (x, y).

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Example of Inverse

Is f : Z2 → Z2 defined by f (x, y) = (x − y, −x) invertible?

Inverse

Now that we have proven bijectivity. Let us find f −1. We know that f (x, y) = (r, s) hence we need to design the inverse, f −1(r, s) = (x, y). Accordin to f , f (x, y) = (x − y, −x) = (r, s). thus x − y = r and −x = s. ⇒ x = −s and y = −s − r.

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Example of Inverse

Is f : Z2 → Z2 defined by f (x, y) = (x − y, −x) invertible?

Inverse

Now that we have proven bijectivity. Let us find f −1. We know that f (x, y) = (r, s) hence we need to design the inverse, f −1(r, s) = (x, y). Accordin to f , f (x, y) = (x − y, −x) = (r, s). thus x − y = r and −x = s. ⇒ x = −s and y = −s − r. Hence, f −1(r, s) = (−s, −s − r).

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Example of Inverse

Is f : Z2 → Z2 defined by f (x, y) = (x − y, −x) invertible?

Inverse

Now that we have proven bijectivity. Let us find f −1. We know that f (x, y) = (r, s) hence we need to design the inverse, f −1(r, s) = (x, y). Accordin to f , f (x, y) = (x − y, −x) = (r, s). thus x − y = r and −x = s. ⇒ x = −s and y = −s − r. Hence, f −1(r, s) = (−s, −s − r). f −1(x − y, −x) = (x, x − x + y) = (x, y).

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Function Composition

Function Composition

Let g be a function from the set A to the set B and let f be a function from the set B to the set C. The composition of the functions f and g, denoted for all a ∈ A by f ◦ g, is defined by (f ◦ g)(a) = f (g(a))

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Function Composition Example

Question

Let f and g be the functions from the set of integers to the set of integers defined by f (x) = 2x + 3 and g(x) = 3x + 2. What is the composition of f and g?

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Function Composition Example

Question

Let f and g be the functions from the set of integers to the set of integers defined by f (x) = 2x + 3 and g(x) = 3x + 2. What is the composition of f and g? (f ◦ g)(x) = f (g(x)) = f (3x + 2) = 2(3x + 2) = 6x + 7.

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Function Composition Example

Let f and g be functions defined as f : Q2 → Q2, g : Q → Q2 such that f (q, r) = (q + r, 2q) and g(x) = (3x, 5 − x).

Does f ◦ g exist?

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Function Composition Example

Let f and g be functions defined as f : Q2 → Q2, g : Q → Q2 such that f (q, r) = (q + r, 2q) and g(x) = (3x, 5 − x).

Does f ◦ g exist?

Yes, f ◦ g : Q → Q2. f ◦ g(x) = f (g(x)).

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Function Composition Example

Let f and g be functions defined as f : Q2 → Q2, g : Q → Q2 such that f (q, r) = (q + r, 2q) and g(x) = (3x, 5 − x).

Does f ◦ g exist?

Yes, f ◦ g : Q → Q2. f ◦ g(x) = f (g(x)). = f (3x, 5 − x)

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Function Composition Example

Let f and g be functions defined as f : Q2 → Q2, g : Q → Q2 such that f (q, r) = (q + r, 2q) and g(x) = (3x, 5 − x).

Does f ◦ g exist?

Yes, f ◦ g : Q → Q2. f ◦ g(x) = f (g(x)). = f (3x, 5 − x)= (3x + 5 − x, 2(3x))

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Function Composition Example

Let f and g be functions defined as f : Q2 → Q2, g : Q → Q2 such that f (q, r) = (q + r, 2q) and g(x) = (3x, 5 − x).

Does f ◦ g exist?

Yes, f ◦ g : Q → Q2. f ◦ g(x) = f (g(x)). = f (3x, 5 − x)= (3x + 5 − x, 2(3x)) = (2x + 5, 6x). f ◦ g(x) = (2x + 5, 6x).

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Function Composition Example

Let f and g be functions defined as f : Q2 → Q2, g : Q → Q2 such that f (q, r) = (q + r, 2q) and g(x) = (3x, 5 − x).

Does f ◦ g exist?

Yes, f ◦ g : Q → Q2. f ◦ g(x) = f (g(x)). = f (3x, 5 − x)= (3x + 5 − x, 2(3x)) = (2x + 5, 6x). f ◦ g(x) = (2x + 5, 6x).

Does g ◦ f exist?

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Function Composition Example

Let f and g be functions defined as f : Q2 → Q2, g : Q → Q2 such that f (q, r) = (q + r, 2q) and g(x) = (3x, 5 − x).

Does f ◦ g exist?

Yes, f ◦ g : Q → Q2. f ◦ g(x) = f (g(x)). = f (3x, 5 − x)= (3x + 5 − x, 2(3x)) = (2x + 5, 6x). f ◦ g(x) = (2x + 5, 6x).

Does g ◦ f exist?

• No. Because co-domain of f is Q2 but the domain of g is Q.

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Composition

Let A, B and C be sets, and let f : A → B and g : B → C be two

• functions. Prove that if f and g are both surjective, then g ◦ f is surjective.

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Composition

Let A, B and C be sets, and let f : A → B and g : B → C be two

• functions. Prove that if f and g are both surjective, then g ◦ f is surjective.

Assume that both f and g are surjective.

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Composition

Let A, B and C be sets, and let f : A → B and g : B → C be two

• functions. Prove that if f and g are both surjective, then g ◦ f is surjective.

Assume that both f and g are surjective. Then,

1 For each b ∈ B there exists an element a ∈ A such that f (a) = b

because f is surjective.

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Composition

Let A, B and C be sets, and let f : A → B and g : B → C be two

• functions. Prove that if f and g are both surjective, then g ◦ f is surjective.

Assume that both f and g are surjective. Then,

1 For each b ∈ B there exists an element a ∈ A such that f (a) = b

because f is surjective.

2 For each c ∈ C there exists an element b ∈ B such that g(b) = c

because g is surjective.

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Composition

Let A, B and C be sets, and let f : A → B and g : B → C be two

• functions. Prove that if f and g are both surjective, then g ◦ f is surjective.

Assume that both f and g are surjective. Then,

1 For each b ∈ B there exists an element a ∈ A such that f (a) = b

because f is surjective.

2 For each c ∈ C there exists an element b ∈ B such that g(b) = c

because g is surjective. Note that g ◦ f : A → C.

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Composition

Let A, B and C be sets, and let f : A → B and g : B → C be two

• functions. Prove that if f and g are both surjective, then g ◦ f is surjective.

Assume that both f and g are surjective. Then,

1 For each b ∈ B there exists an element a ∈ A such that f (a) = b

because f is surjective.

2 For each c ∈ C there exists an element b ∈ B such that g(b) = c

because g is surjective. Note that g ◦ f : A → C. Now, take c ∈ C, then g ◦ f (a) = g(f (a))

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Composition

Let A, B and C be sets, and let f : A → B and g : B → C be two

• functions. Prove that if f and g are both surjective, then g ◦ f is surjective.

Assume that both f and g are surjective. Then,

1 For each b ∈ B there exists an element a ∈ A such that f (a) = b

because f is surjective.

2 For each c ∈ C there exists an element b ∈ B such that g(b) = c

because g is surjective. Note that g ◦ f : A → C. Now, take c ∈ C, then g ◦ f (a) = g(f (a))= g(b) (by (1))

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Composition

Let A, B and C be sets, and let f : A → B and g : B → C be two

• functions. Prove that if f and g are both surjective, then g ◦ f is surjective.

Assume that both f and g are surjective. Then,

1 For each b ∈ B there exists an element a ∈ A such that f (a) = b

because f is surjective.

2 For each c ∈ C there exists an element b ∈ B such that g(b) = c

because g is surjective. Note that g ◦ f : A → C. Now, take c ∈ C, then g ◦ f (a) = g(f (a))= g(b) (by (1)) = c (by (2)). Thus, g ◦ f is surjective.

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Composition

Show that if g ◦ f is surjective then g is surjective.

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Composition

Show that if g ◦ f is surjective then g is surjective. Suppose that g ◦ f is surjective.

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Composition

Show that if g ◦ f is surjective then g is surjective. Suppose that g ◦ f is surjective. Let z ∈ C. Then since g ◦ f is surjective, there exists x ∈ A such that (g ◦ f )(x) = g(f (x)) = z.

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Composition

Show that if g ◦ f is surjective then g is surjective. Suppose that g ◦ f is surjective. Let z ∈ C. Then since g ◦ f is surjective, there exists x ∈ A such that (g ◦ f )(x) = g(f (x)) = z. Therefore if we let y = f (x) ∈ B, then g(y) = z.

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Composition

Show that if g ◦ f is surjective then g is surjective. Suppose that g ◦ f is surjective. Let z ∈ C. Then since g ◦ f is surjective, there exists x ∈ A such that (g ◦ f )(x) = g(f (x)) = z. Therefore if we let y = f (x) ∈ B, then g(y) = z. Thus g is surjective.

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Composition

In each part of the exercise, give examples of sets A, B, C and functions f : A → B and g : B → C satisfying the indicated properties. (a) g is not injective but g ◦ f is injective. (b) f is not surjective but g ◦ f is surjective.

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Composition

In each part of the exercise, give examples of sets A, B, C and functions f : A → B and g : B → C satisfying the indicated properties. (a) g is not injective but g ◦ f is injective. (b) f is not surjective but g ◦ f is surjective. The same example works for both. Let A = {1}, B = {1, 2}, C = {1}, and f : A → B by f (1) = 1 and g : B → C by g(1) = g(2) = 1.

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Composition

In each part of the exercise, give examples of sets A, B, C and functions f : A → B and g : B → C satisfying the indicated properties. (a) g is not injective but g ◦ f is injective. (b) f is not surjective but g ◦ f is surjective. The same example works for both. Let A = {1}, B = {1, 2}, C = {1}, and f : A → B by f (1) = 1 and g : B → C by g(1) = g(2) = 1. Then g ◦ f : A → C is defined by (g ◦ f )(1) = 1.

14/15

Composition

In each part of the exercise, give examples of sets A, B, C and functions f : A → B and g : B → C satisfying the indicated properties. (a) g is not injective but g ◦ f is injective. (b) f is not surjective but g ◦ f is surjective. The same example works for both. Let A = {1}, B = {1, 2}, C = {1}, and f : A → B by f (1) = 1 and g : B → C by g(1) = g(2) = 1. Then g ◦ f : A → C is defined by (g ◦ f )(1) = 1. This map is a bijection from A = {1} to C = {1}, so is injective and surjective.

14/15

Composition

In each part of the exercise, give examples of sets A, B, C and functions f : A → B and g : B → C satisfying the indicated properties. (a) g is not injective but g ◦ f is injective. (b) f is not surjective but g ◦ f is surjective. The same example works for both. Let A = {1}, B = {1, 2}, C = {1}, and f : A → B by f (1) = 1 and g : B → C by g(1) = g(2) = 1. Then g ◦ f : A → C is defined by (g ◦ f )(1) = 1. This map is a bijection from A = {1} to C = {1}, so is injective and surjective. However, g is not injective, since g(1) = g(2) = 1, and f is not surjective, since 2 / ∈ f (A) = 1.

14/15

Composition

Define functions f and g from Z to Z such that f is not surjective and yet g ◦ f is surjective.

15/15

Composition

Define functions f and g from Z to Z such that f is not surjective and yet g ◦ f is surjective. Let, The map f is not surjective: the image is the set of even integers. However, g ◦ f is surjective, since (g ◦ f )(n) = g(f (n)) = g(2n)) = (2n)/2 = n so in fact g ◦ f is the identity map on Z.

15/15

Thank you for your attention!!