Suppose that A,B,C,D are sets with A≈C and B≈D. Prove that AxB≈DxC?

we have bijections f from A to C and g from B to D by definition of "≈". define a map h from AxB to DxC by h(a,b) = (g(b),f(a)). we have to check that h is injective and surjective to show that it is bijective. for injectivity, suppose h(a,b)=h(c,d). then (g(b),f(a)) = (g(d),f(c)), which means g(b)=g(d) and f(a)=f(c). since f and g are injective, this means b=d and a=c, so (a,b)=(c,d). thus h is injective. to check that h is surjective, suppose (d,c) is in DxC. since g is surjective, there is some b in B such that g(b)=d. since f is surjective, there is some a in A such that f(a)=c. now, h(a,b) = (g(b),f(a)) = (d,c), so h is surjective. thus h is bijective and AxB is in bijection with DxC.

Suppose that A,B,C,D are sets with A≈C and B≈D. Prove that AxB≈DxC?
Is this commutative DxC = CxD ?

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