12 1.2 Functions Definition 1.2.2 Let f : X →Y be a function. (i) We say f is surjective (or maps X onto Y) if for every element y ∈Y, there exists an element x ∈X such that f (x) =y. (ii) We say f is injective (or one-to-one) if whenever x1,x2 ∈X, x1̸ =x2, then f (x1)̸ = f (x2). Equivalently, f is one-to-one if for all x1,x2 ∈X with f (x1) =f (x2), it follows that x1 =x2. (iii) If f is both surjective and injective, we say f is bijective (or a one-to-one correspondence). Remark 1.2.1 When the function f is a bijection, for anyy∈Y, there exists a unique element x ∈X such that f (x) =y. This element x is then denoted by f −1(y). In this way, we already built a function fromY toX called the inverse of f . Theorem 1.2.1 Let f : X →Y. If there are two functions g: Y →X and h: Y →X such that g(f (x)) =x for every x ∈X and f (h(y)) =y for every y ∈Y, then f is bijective and g=h=f −1. Proof: First we prove that f is surjective. Let y ∈Y and set x =h(y). Then, from the assumption on h, we have f (x) =f (h(y)) =y. This shows that f is surjective. Next we prove that f is injective. Let x1,x2 ∈X be such that f (x1) = f (x2). Then x1 = g(f (x1)) =g(f (x2)) =x2. Thus, f is injective. We have shown that for each y ∈Y, there is a unique x ∈X, which we denote f −1(y), such that f (x) =y. Since for such a y, g(y) =g(f (x)) =x, we obtaing(y) =f −1(y). Since f (h(y)) =y, we also conclude that h(y) =x =f −1(y). □ ■Example 1.2.1 Consider the function f : (1,2] →[3,4) given by f (x) =4−(x−1) 2. We show that f is bijective. First let x1,x2 ∈(1,2] be such that f (x1) =f (x2). That is, 4−(x1−1) 2 =4−(x 2−1) 2. Then (x1 −1) 2 = (x 2 −1) 2. Since bothx 1 >1 and x2 >1, we conclude that x1 −1=x2 −1 and, so, x1 =x2. This proves f is injective. Next let y∈[3,4). We need to findx∈(1,2] such that f (x) =y. For that, we set up 4−(x−1)2 = y and solve for x. We get, x =√4−y+1. Note that since y <4, 4−y has a square root. Also note that since 3≤y <4, we have 1≥4−y >0 and, hence, 2≥ √4−y+1>1. Therefore, x ∈(1,2]. This proves f is surjective. Definition 1.2.3 Let f : X →Y be a function and let Abe a subset of X. Then the image of A under f is given by f (A) ={f (a) : a∈A}. It follows from the definition that f (A) ={b∈Y : b= f (a) for some a∈A}. Moreover, f is surjective if and only if f (X) =Y. Definition 1.2.4 Let f : X →Y be a function and let Bbe a subset of Y. Then the preimage of B under f is given by f −1(B) ={x ∈X : f (x) ∈B}. Remark 1.2.2 Note that, despite the notation, the definition of preimage does not require the function to have an inverse. It does not even require the function to be injective. The examples below illustrate these concepts.
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