Let A and B be sets. Show that f:A×B→B×A such that f(a,b)=(b,a) is bijective function.
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Solution
f:A×B→B×A is defined as f(a,b)=(b,a).
Let (a1,b1),(a2,b2)∈A×B such that f(a1,b1)=f(a2,b2). ⇒(b1,a1)=(b2,a2) ⇒b1=b2) and (a1=a2) ⇒(a1,b1)=(a2,b2) ∴f is one-one. Now, let (b,a)∈B×A be any element. Then, there exists (a,b)∈A×B such that