Proof involving functions and intersection

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Prove that f(\cap T_\alpha)=\cap f(T_\alpha) for all choices of (T_\alpha) \alpha \in \lambda if and only if f is one-to-one.

I've been working on this on and off for a day and have nothing to show for it... Any help pointing me in the right direction would be appreciated.

Also, more important then this question, can any of you recommend additional resources to review this type of stuff, mostly functions, order isomorphisms, etc. The book I am using does not give as many examples as I'd like.
 
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It's not really that hard. First assume f is one-to-one. Now can you show that if y is an element of f(\cap T_\alpha) then it is an element of \cap f(T_\alpha) and vice versa? For the opposite direction I'd do the contrapositive. Show if f is not one-to-one then you can find a counterexample.
 
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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