"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}.