Inverses and Composition for Relations

Definition: If R is relation R from A to B, then R−1 is a relation from B to A defined by R−1 = {(b,a) : (a,b)∈ R} . R−1 is called the inverse of R. Proposition: (i) (R−1)−1 = R. (ii) Dom( R−1 ) = Ran(R) and Dom(R) = Ran( R−1 ). Proof: (i) (a,b)∈ R if and only if (b,a)∈ R−1 , and (b,a)∈ R−1 if and only if (a,b)∈ (R−1)−1 . Therefore R and (R−1)−1 contain the same ordered pairs, and hence they are equal. Let b∈ Dom(R−1) . Then there exists a ∈ A such that (b,a)∈ R−1 . Therefore (a,b)∈ R , and so b∈ Ran(R) . Now let b∈ Ran(R) . Then there exists a ∈ A such that (a,b)∈ R . Then (b,a)∈ R−1 , and so b∈ Dom(R−1) . Therefore Dom(R−1 ) = Ran(R). Now replace R by R−1 in that statement. Then Dom( (R−1)−1 ) = Ran( R−1 ), and since (R−1)−1 = R, Dom(R) = Ran(R−1 ). Definition: For any set A, the relation IA = {(x, x) : x ∈ A} is called the identity relation on A. Definition: If R is relation R from A to B, and S is a relation from B to C, then S R is a relation from A to C defined by S R = { (a,c)∈ A×C : ∃ b∈ B such that (a,b)∈ R and (b,c)∈ S }. S R is called the composition of R and S. Proposition: Let R be a relation from A to B, S be a relation from B to C, and T be a relation from C to D. (i) IB R = R and R IA = R (ii) (S R)−1 = R−1 S−1 (iii) T (S R) = (T S)R

EXERCISES R1. Let R = {(1,5), (2,2), (3,4), (5,2)}, S = {(2,4), (3,4), (3,1), (5,5)}, and T = {(1,4), (3,5), (4,1)}. Find (a) R S (b) R T (c) T S (d) R R (e) S R (f) T T (g) R (S T ) (h) (R S)T R2. Let A = {1, 2, 3}. List the ordered pairs and draw the directed graph of a relation on A with the given properties. (a) not reflexive, not symmetric, not transitive (b) reflexive, not symmetric, not transitive (c) not reflexive, symmetric, not transitive (d) reflexive, symmetric, not transitive (e) not reflexive, not symmetric, transitive (f) reflexive, not symmetric, transitive (g) not reflexive, symmetric, transitive (h) reflexive, symmetric, transitive R3. Prove or disprove. (a) If R is a relation from A to B, then R−1 R = IA and R R−1 = IB . (b) If R is a relation on A, then R∪R−1 is symmetric.

One-to-one and Onto Relations Definition: A relation R from A to B is one-to-one if whenever

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(x,z)∈ R and

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(y,z)∈ R ,

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x = y . Definition: A relation R from A to B is onto if Ran(R) = B, in other words for all

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y ∈ B, there exists

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x ∈ A so that

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(x,y)∈ R. In a sense the definition of “one-to-one” is the inverse of condition (ii) of the definition of a function. And the definition of “onto” is the inverse of condition (i) of the definition of a function (that Dom(f) = A). This is made explicit in Theorem A. Theorem A: Let R be a relation from A to B. (i) R is one-to-one if and only if

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R−1 is a function from Ran(R) to A. (ii) R is onto if and only if Dom(

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R−1) = B. Proof: (i) Assume R is one-to-one, and suppose

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(x,y)∈ R−1 and

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(x,z)∈ R−1. Then

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(y,x)∈ R and

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(z,x)∈ R . Since R is one-to-one,

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y = z. Therefore

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R−1 meets condition (ii) of the definition of a function. Since

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R−1 is a relation from B to A, and since

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Dom(R−1) = Rng(R) by Theorem 3.2(b),

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R−1 is a function from Ran(R) to A. Now assume

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R−1 is a function from Ran(R) to A, and let

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(x,z)∈ R and

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(y,z)∈ R . Then

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z ∈ Rng(R) =Dom(R−1) , and

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(z,x)∈ R−1 and

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(z,y)∈ R−1. Since

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R−1 is a function,

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x = y . Hence R is one-to-one. (ii) By definition, R is onto if and only if Ran(R) = B. By Theorem 3.2(b) Ran(R) = B if and only if Dom(

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R−1) = B. ❚ If f is a function from A to B, then

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f −1 always exists as a relation, and consists of the reverses of all the ordered pairs in f. But

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f −1 may not be a function. Corollary B: If f is a function from A to B, then

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f −1 is a function from Ran(f) to A if and only if f is one-to-one. Proof: This is just part (i) of Theorem A. ❚ Theorem C: f is a one-to-one and onto function from A to B if and only if

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f −1 is a one-to-one and onto function from B to A. Proof: From Theorem A(i),

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f −1 is a function if and only if f is one-to-one. Replacing f by

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f −1, we see that f is a function if and only if

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f −1 is one-to-one. Therefore f is a one-to-one function if and only if

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f −1 is a one-to-one function. From Theorem A(ii). we see that f is onto if and only if Dom(

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f −1) = B. Replacing f by

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f −1, we see that

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f −1 is onto if and only if Dom(f) = A. Now suppose that f is a one-to-one and onto function from A to B. Then we have shown that

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f −1 is a one-to-one function, Dom(

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f −1) = B, and

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f −1 is onto. Conversely if

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f −1 is a one-to-one and onto function from B to A, we have shown that f is a one-to-one function, Dom(f) = A, and f is onto. ❚