Don't over think it.
There is a bijection .
Hi, so I'm suuposed to prove the properties of a equivalence relation. I have proved the reflexive and transitive property but I'm confusing myself with the transitive property so far I have assumed that A ~ B and i want to show B~A. so i started by showing that a function from B to A is one to one so I said let f inverse go from B to A. I'm not sure if I'm allowed to let it be f inverse.
then because A~B we know that f(a) = b so
f-1(b1) = f-1(b2) becomes f-1(f(a1)) = f-1(f(a2)) then I canceled the functions which I'm also not sure whether its alright to do that and got a1 = a2 then because A~B its injective so if a1 = a2 then f(a1) = f(a2). Im not sure whether I'm allowed to do that either. then i proceed to once again say because A~B its surjective so if f(a1)=f(a2) then b1 = b2
I'm also very confused with the surjective part if anyone can help. Thanks!!