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We know that f:A~B is a bijection
Therefore f^-1:A~B is a unique function
To prove that f^-1 is one-one--
Let b1, b2 be any 2 different elements of B ,, i.e
b1 is unequal to b2
Now we have to prove that
f^-1(b1) is unequal to f^-1(b2)
Let f^-1(b1)=a1. And. f^-1(b2)=a2
Such that a1,a2 €A
Then b1= f(a1) and b2=f(a2)
~f^-1(b1) is unequal to f^-1(b2)
Therefore f^-1 is a one-one function
Now f^-1 has a n image a such that
b€B
Therefore f^-1 is onto function
Finally f^-1 is a bijection
Hence proved
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