(1) f : A → B iff ∀x ∈ A , ∃! y ∈ B and y = f(x). f(A) ⊂ B or the most is equal to B, so f(A) ⊆ B .
if f(x) ∈ f(A) , then x ∈ A . For example , A = { -2,1,2 } , f(x) = x^2 , f(A) = { 1,4} . Assume f(x) = 1 , f(x) ∈ f(A) , but -1 ∉ A .
if x ∈ A , f(x) ∈ f(A) ∩ B . Basically , f(A) ⊆ B . f(A) ∩ B = f(A)
D ⊂ A , then f(D) ⊂ B . For example , f(x) = x^2 , A = {1,2,3}, D = {1,2} , f(D) = { 1,4 } , B = { 1,4,9,5,6,7 } To prove this fact , please see below :
For any element x , x ∈ D and D ⊂ A , f(x) ∈ f(D) ⊂ f(A) ⊂ B . Therefore, f(D) ⊂ B .
onto function : For any element y , y ∈ B , there exists an element x (not only one ) , x ∈ A, such that y = f(x). 也就是說, B中所有的elements都被A中的元素對應via f(x) . 若 B ⊂ f(A) , 則 B中的元素絕對被f(x)所映射到. For example , A = {-1,1,2,3} , f(x) = x^2 , B = {1,4} , f(A) = {1,4,9}
(2) C = { -1,0,1,2 } , f: C → Z f(x) = 2x , x < 0
1 , x = 0
x+3 , x > 0
C 就是那個 domain , C中所有的元素藉由f全部被映射到Z中,
f(-1) = -2 , f(0) = 1 , f(1) = 4 , f(2) = 5 所以f的值域為 {-2,1,4,5}
f(C) ⊂ Z .



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