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# Pick any y b there exists someσ i such that y b σ

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Pick any y B . There exists some σ I such that y B σ . For the same σ , since f σ is surjective, there exists some x A σ such that y = f σ ( x ). Therefore, by the definition of the function f , we have y = f ( x ) for the same x . It follows that f is a surjective function. (c) The following statement is false in general: ‘Suppose f α is an injective function for any α I . Then f is injective.’ We illustrate this by constructing an appropriate counter-example here: Let I = { 0 , 1 } , where 0 , 1 are regarded as distinct objects. Let A 0 = { 2 } , A 1 = { 3 } , where 2 , 3 are regarded as distinct objects. We have A = A 0 A 1 = { 2 , 3 } . Let B 0 = B 1 = { 4 } . We have B = B 0 B 1 = { 4 } Define the function f 0 : A 0 −→ B 0 by f 0 (2) = 4. Define the function f 1 : A 1 −→ B 1 by f 1 (3) = 4. f 0 ,f 1 are both injective. Hence f α is an injective function for any α I . We have A 0 A 0 = { 2 } , A 1 A 1 = { 3 } and A 0 A 1 = A 1 A 0 = . Hence for any α,β I , for any x A α A β , we have f α ( x ) = f β ( x ). The function f : A −→ B is given by f (2) = f (3) = 4. f is not an injective function. 8

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