Introduction to Mathematical Analysis I - Second Edition

12 1.2 FUNCTIONS 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 x , x 0 ∈ X be such that f ( x ) = f ( x 0 ) . Then x = g ( f ( x )) = g ( f ( x 0 )) = x 0 . 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 obtain g ( 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 x , y ∈ ( 1 , 2 ] be such that f ( x ) = f ( y ) . That is, 4 − ( x − 1 ) 2 = 4 − ( y − 1 ) 2 . Then ( x − 1 ) 2 = ( y − 1 ) 2 . Since both x > 1 and y > 1, we conclude that x − 1 = y − 1 and, so, x = y . This proves f is injective. Next let y ∈ [ 3 , 4 ) . We want x ∈ ( 1 , 2 ] such that f ( x ) = y . Let us set up 4 − ( x − 1 ) 2 = y and solve for x . We get, x = √ 4 − y + 1. Note that since y < 4, y − 4 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 A be 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 . For a subset B of Y , the preimage of B under f is defined 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. Example 1.2.2 Let f : R → R be given by f ( x ) = 3 x − 1. Let A = [ 0 , 2 ) and B = { 1 , − 4 , 5 } . Then f ( A ) = [ − 1 , 5 ) and f − 1 ( B ) = { 2 3 , − 1 , 2 } . Example 1.2.3 Let f : R → R be given by f ( x ) = − x + 7. Let A = [ 0 , 2 ) and B = ( − ∞ , 3 ] . Then f ( A ) = ( 5 , 7 ] and f − 1 ( B ) = [ 4 , ∞ ) . Example 1.2.4 Let f : R → R be given by f ( x ) = x 2 . Let A = ( − 1 , 2 ) and B = [ 1 , 4 ) . Then f ( A ) = [ 0 , 4 ) and f − 1 ( B ) = ( − 2 , − 1 ] ∪ [ 1 , 2 ) . Theorem 1.2.3 Let f : X → Y be a function, let A be a subset of X , and let B be a subset of Y . The following hold: (a) A ⊂ f − 1 ( f ( A )) .