Set theory, functions, bijectivity

Daveyboy
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f:X\rightarrowY, A\subsetX
f(Ac)=[f(A)]c implies f bijective.

Just trying to apply the definitions of injective and bijective. The equivalence makes sense but I am having a hard time showing it.

f(x)=f(y) implies x=y and for every y in Y there exists a x in X s.t. f(x)=y.

I mean all I have is if f(x)=f(y)
then f(X\x)=Y\f(x\x)=f(X\y)-Y-f(X\y)...
 
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what's the question?
 
Daveyboy said:
f(Ac)=[f(A)]c implies f bijective.
this implication
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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