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Image of Intersection equals Intersection of the Images iff Function is injective.

  • #1

Homework Statement



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

Homework Equations



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

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 [tex]b\in B[/tex].
 

Answers and Replies

  • #2
Dick
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Homework Helper
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Try a proof by contradiction. Assume f is NOT injective. Can you construct two sets that violate the equality?
 
  • #3


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 [tex]f[/tex] is not inyective. Then there exists [tex]b\in \Im f[/tex] such that there are at least two [tex]x_{1},x_{2}\in A[/tex] such that are preimages of [tex]b[/tex] vía [tex]f[/tex].

Let [tex]C[/tex] be the set of preimages of [tex]b[/tex] with the exception of only one, and let [tex]D[/tex] the set having the one missing in [tex]C[/tex]. Then [tex]C\cap D=\emptyset[/tex] and [tex]f(C)=f(D)=\{b\}\rightarrow f(C)\cap f(D)=\{b\}[/tex]. But we have supposed that [tex]f(C\cap D)=f(C)\cap f(D)[/tex], however the contention [tex]\emptyset\supset\{b\}[/tex] is false by definition of empty set. Therefore [tex]f[/tex] is inyective.
 
  • #4
Dick
Science Advisor
Homework Helper
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That looks ok. It might be a little clearer if you just say C={x1} and D={x2}.
 

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