A/Q the functions f:A → B and g:C → D are bijections Then g-1 must exist . Then for a function h∈ Ac we may define a function T: AC → BBD by T(h) = f o h o g-1 . That is , for d∈D, T(h)(d) = f(h(g-1(d))) Since g-1:D→C, the expression g-1(d) must exist Now, As h:C → A and g-1(d)∈C then the expression h(g-1(d)) must exist Again, As h(g-1(d)) ∈ A and f:A → B, the expression f(h(g-1(d))) must exist Now it only remains to prove that R(h) = f o h o g-1 is a bijection. To do so, we need to simply provide an inverse. Now T o R(h) = f o (f-1 o h o g) o g-1 = (f o f-1) o h o (g o g-1 ) = idB o h o idD = h Therefore R:h → f-1 oh o g exists and is an inverse to T Hence there exists a bijection between AC + BD Correct Answer: Yes.
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