It follows that f P is surjective iv We are going to prove that F implies D

It follows that f p is surjective iv we are going to

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It follows that f P is surjective. iv. We are going to prove that (F) implies (D): Suppose f P is surjective. Pick any x,y A . Suppose f ( x ) = f ( y ). We have { x } , { y } ∈ P ( A ). Since f P is surjective, there exist some U,V P ( B ) such that { x } = f P ( U ) and { y } = f P ( V ). Then { x } = f 1 ( U ) and { y } = f 1 ( V ). Since { y } = f 1 ( V ), we have y f 1 ( V ). Then f ( y ) V . Since f ( x ) = f ( y ), we have f ( x ) V as well. Then x f 1 ( V ). Since x ∈ { x } = f 1 ( U ), we have x f 1 ( U ) f 1 ( V ) = { x } ∩ { y } . Therefore x = y . (If it were true that x negationslash = y then { x } ∩ { y } = .) It follows that f is injective. 5. Let A,B,C,D be non-empty sets. Suppose there is a bijective function from A to C and there is a bijective function from B to D . Pick a bijective function α : A −→ C and a bijective function β : B −→ D indeed. 5
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(a) Let G be the subset of ( A × B ) × ( C × D ) given by G = { ( s,t ) | s = ( x,y ) and t = ( α ( x ) ( y )) for some x A,y B } , and γ = ( A × B,C × D,G ). We verify that γ is a bijective function: Pick any s A × B . By definition s = ( x,y ) for some x A,y B . We have α ( x ) C , β ( y ) D , ( α ( x ) ( y )) C × D . Write t = ( α ( x ) ( y )). We have ( s,t ) = (( x,y ) , ( α ( x ) ( y ))) G . Pick any s A × B and t,t C × D . Suppose ( s,t ) , ( s,t ) G . Since ( s,t ) G , we have s = ( x,y ) ,t = ( α ( x ) ( y )) for some x A,y B . Since ( s ,t ) G , we have s = ( x ,y ) ,t = ( α ( x ) ( y )) for some x A,y B . We have ( x,y ) = s = ( x ,y ) (with x,x A and y,y B ). Then x = x and y = y . Therefore α ( x ) = α ( x ), β ( y ) = β ( y ), and t = ( α ( x ) ( y )) = ( α ( x ) ( y )) = t . Pick any t C × D . Then t = ( u,v ) for some u C,v D . Since α is a surjective function, u = α ( x ) for some x A . Since β is a surjective function, v = β ( y ) for some y B . We have ( x,y ) A × B . Write s = ( x,y ). We have ( s,t ) = (( x,y ) , ( α ( x ) ( y ))) G . Pick any t C × D and any s,s A × B . Suppose ( s,t ) , ( s ,t ) G . Since ( s,t ) G , we have s = ( x,y ) ,t = ( α ( x ) ( y )) for some x A,y B . Since ( s ,t ) G , we have s = ( x ,y ) ,t = ( α ( x ) ( y )) for some x A,y B . We have ( α ( x ) ( y )) = t = ( α ( x ) ( y )) (with α ( x ) ( x ) C , β ( y ) ( y ) D ). Then α ( x ) = α ( x ) and β ( y ) = β ( y ). Since α is an injective function, we have x = x . Since β is an injective function, we have y = y . Therefore s = ( x,y ) = ( x ,y ) = s . We have verified that γ = ( A × B,C × D,G ) is a bijective function. (The function γ is given ‘explicitly’ by γ ( x,y ) = ( α ( x ) ( y )) for any x A,y B .) (b) We make some observations first: Since α is a bijective function from A to C , α 1 is a bijective function from C to A . Since β is a bijective function from B to D , β 1 is a bijective function from D to B . If ϕ is a function from A to B then β ϕ α 1 is a function from C to D . If ψ is a function from C to D then β 1 ψ α is a function from A to B . Let H be the subset of Map ( A,B ) × Map ( C,D ) given by H = { ( ϕ,β ϕ α 1 ) | ϕ Map ( A,B ) } , and Γ = ( Map ( A,B ) , Map ( C,D ) ,H ). By definition, Γ is a function. (The function Γ is given ‘explicitly’ by Γ( ϕ ) = β ϕ α 1 for any ϕ Map ( A,B ).) We verify that Γ is a bijective function: Pick any ψ Map ( C,D ). We have β 1 ψ α Map ( A,B ). Moreover, Γ( β 1 ψ α ) = β ( β 1 ψ α ) α 1 = ( β β 1 ) ψ ( α α 1 ) = id D ψ id C = ψ. Pick any ϕ,ϕ Map ( C,D ). Suppose Γ( ϕ ) = Γ( ϕ ). Then we have ϕ = β 1 β ϕ α 1 α = β 1 Γ( ϕ ) α = β 1 Γ( ϕ ) α = β 1 β ϕ α 1 α = ϕ .
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