MAT246H1-S – LEC0201/9201
Concepts in Abstract Mathematics
REVIEWS ABOUT FUNCTIONS
March 23rd, 2021
Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Mar 23, 2021 1 / 6
Functions – 1
(Informal) definition of a function
Afunction(ormap) is the data of two sets𝐴and𝐵together with a ”process” which assigns to each𝑥 ∈ 𝐴a unique𝑓 (𝑥) ∈ 𝐵:
𝑓 ∶{
𝐴 → 𝐵
𝑥 ↦ 𝑓 (𝑥)
Here,𝑓 is the name of the function,𝐴is thedomainof𝑓, and𝐵is thecodomainof𝑓.
Remark
This process can be:
• A formula:𝑓 ∶ ℝ → ℝdefined by𝑓 (𝑥) = 𝑒𝑥2−𝜋+ 42.
• An exhaustive list:𝑓 ∶ {1, 2, 3} → ℝdefined by𝑓 (1) = 𝜋,𝑓 (2) = √2,𝑓 (3) = 𝑒.
• A property characterizing𝑓:logis the unique antiderivative of𝑔 ∶ (0, +∞) → ℝdefined by𝑔(𝑥) = 1 such thatlog(1) = 0. 𝑥
• By induction: we define the sequence𝑢𝑛∶ ℕ → ℝby𝑢0= 1and∀𝑛 ∈ ℕ, 𝑢𝑛+1= 𝑢2𝑛+ 1.
• …
Functions – 2
Remark
The domain and codomain are part of the definition of a function.
For instance:
• 𝑓 ∶ {
ℝ → (0, +∞)
𝑥 ↦ 𝑒𝑥 and 𝑔 ∶
{
ℝ → ℝ
𝑥 ↦ 𝑒𝑥
are not the same function (the first one is surjective but not the second one).
• 𝑓 ∶ {
[0, +∞) → ℝ
𝑥 ↦ 𝑥2+ 1 and 𝑔 ∶
{
ℝ → ℝ
𝑥 ↦ 𝑥2+ 1
are not the same function (the first one is injective but not the second one).
A function is not simply a ”formula”, you need to specify the domain and the codomain.
Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Mar 23, 2021 3 / 6
Injective/Surjective/Bijective functions – 1
Injective/Surjective/Bijective functions Given a function𝑓 ∶ 𝐴 → 𝐵.
• We say that𝑓 isinjective(orone-to-one) if ∀𝑥1, 𝑥2∈ 𝐴, 𝑥1≠ 𝑥2 ⟹ 𝑓 (𝑥1) ≠ 𝑓 (𝑥2) or equivalently by taking the contrapositive ∀𝑥1, 𝑥2∈ 𝐴, 𝑓 (𝑥1) = 𝑓 (𝑥2) ⟹ 𝑥1= 𝑥2
• We say that𝑓 issurjective(oronto) if ∀𝑦 ∈ 𝐵, ∃𝑥 ∈ 𝐴, 𝑦 = 𝑓 (𝑥)
• We say that𝑓 isbijectiveif it is injective and surjective, i.e. ∀𝑦 ∈ 𝐵, ∃!𝑥 ∈ 𝐴, 𝑦 = 𝑓 (𝑥) 𝑎
𝑏 𝑐
1 2 3 4
Figure:Injective
𝑎 𝑏 𝑐 𝑑
1 2 3
Figure:Surjective
𝑎 𝑏 𝑐 𝑑
1 2 3 4
Figure:Bijective
𝑎 𝑏 𝑐
1 2 3 4
Figure:Not injective
𝑎 𝑏 𝑐 𝑑
1 2 3 4
Figure:Not surjective
Injective/Surjective/Bijective functions – 2
Proposition
Let𝑓 ∶ 𝐸 → 𝐹 and𝑔 ∶ 𝐹 → 𝐺be two functions.
1 If𝑓 and𝑔are injective then so is𝑔 ∘ 𝑓.
2 If𝑓 and𝑔are surjective then so is𝑔 ∘ 𝑓.
3 If𝑔 ∘ 𝑓 is injective then𝑓 is injective too.
4 If𝑔 ∘ 𝑓 is surjective then𝑔is surjective too.
Proof.
1 Let𝑥, 𝑦 ∈ 𝐸be such that𝑔(𝑓 (𝑥)) = 𝑔(𝑓 (𝑦)).
Then𝑓 (𝑥) = 𝑓 (𝑦)since𝑔is injective. Thus𝑥 = 𝑦since𝑓 is injective.
2 Let𝑧 ∈ 𝐺. Since𝑔 is surjective, it exists𝑦 ∈ 𝐹 such that𝑧 = 𝑔(𝑦).
Since𝑓 is surjective, it exists𝑥 ∈ 𝐸such that𝑦 = 𝑓 (𝑥). Therefore𝑧 = 𝑔(𝑓 (𝑥)).
3 Let𝑥, 𝑦 ∈ 𝐸such that𝑓 (𝑥) = 𝑓 (𝑦).
Then𝑔(𝑓 (𝑥)) = 𝑔(𝑓 (𝑦))and thus𝑥 = 𝑦since𝑔 ∘ 𝑓 is injective.
4 Let𝑧 ∈ 𝐺. Since𝑔 ∘ 𝑓 is surjective, there exists𝑥 ∈ 𝐸 such that𝑧 = 𝑔(𝑓 (𝑥)).
Then𝑦 = 𝑓 (𝑥) ∈ 𝐹 satisfies𝑔(𝑦) = 𝑧. ■
Jean-Baptiste Campesato MAT246H1-S – LEC0201/9201 – Mar 23, 2021 5 / 6
Inverse of a bijection
Proposition
𝑓 ∶ 𝐴 → 𝐵is bijective if and only if there exists𝑔 ∶ 𝐵 → 𝐴such that {
∀𝑥 ∈ 𝐴, 𝑔(𝑓 (𝑥)) = 𝑥
∀𝑦 ∈ 𝐵, 𝑓 (𝑔(𝑦)) = 𝑦 . Then𝑔 is unique, it is called theinverse of𝑓 and denoted by𝑓−1 ∶ 𝐵 → 𝐴.
Proof.
⇒Assume that𝑓 is bijective, then∀𝑦 ∈ 𝐵, ∃!𝑥𝑦∈ 𝐴, 𝑓 (𝑥𝑦) = 𝑦.
We define𝑔 ∶ 𝐵 → 𝐴by𝑔(𝑦) = 𝑥𝑦. Then𝑔satisfies the required properties.
⇐Assume that there exists𝑔as in the statement.
Then𝑔 ∘ 𝑓 = 𝑖𝑑𝐴is injective, so𝑓is too.
And𝑓 ∘ 𝑔 = 𝑖𝑑𝐵is surjective, thus𝑓 is too.
Therefore𝑓 is bijective.
Uniqueness:assume there exist two such functions𝑔1, 𝑔2∶ 𝐵 → 𝐴.
Let𝑦 ∈ 𝐵. Then𝑓 (𝑔1(𝑦)) = 𝑦 = 𝑓 (𝑔2(𝑦)).
So𝑔1(𝑦) = 𝑔2(𝑦)since𝑓 is injective. ■
𝑎 𝑏 𝑐 𝑑
1 2 3 4
Figure:Bijective function
𝑎 𝑏 𝑐 𝑑
1 2 3 4
Figure:Its inverse
