"Intersection Equality iff Function is Injective

[ELESSAR TELKONT]

Homework Statement

Let A, B be sets, C,D\subset A and f:A\longrightarrow B be a function between them. Then f(C\cap D)=f(C)\cap f(D) if and only if f is injective.

Homework Equations

Another proposition, that I have proven that for any function f(C\cap D)\subset f(C)\cap f(D), and the definition of injectiveness: f is inyective if \forall b\in B\mid b=f(x)=f(y) for some x,y\in A implies that x=y.

The Attempt at a Solution

If we suppose the injectiveness is trivial to get the equality. But for the other direction I get stuck in what way to use the equality of images to get inyection. I can't see how to make a proof, in fact I can't associate the equality with the fact that there must be a unique preimage for every b\in B.

 

[Dick]

Try a proof by contradiction. Assume f is NOT injective. Can you construct two sets that violate the equality?

 

[ELESSAR TELKONT]

Thanks. I have not thought in that manner previously. Here is my proof of the implication I had problems with:

Let the equality is true. Suppose that f is not inyective. Then there exists b\in \Im f such that there are at least two x_{1},x_{2}\in A such that are preimages of b vía f.

Let C be the set of preimages of b with the exception of only one, and let D the set having the one missing in C. Then C\cap D=\emptyset and f(C)=f(D)=\{b\}\rightarrow f(C)\cap f(D)=\{b\}. But we have supposed that f(C\cap D)=f(C)\cap f(D), however the contention \emptyset\supset\{b\} is false by definition of empty set. Therefore f is inyective.

 

[Dick]

That looks ok. It might be a little clearer if you just say C={x1} and D={x2}.