"Intersection Equality iff Function is Injective

In summary, for sets A, B and subsets C, D of A, and a function f from A to B, the equality f(C\cap D)=f(C)\cap f(D) holds if and only if f is injective. To prove the implication from the equality to injectiveness, a proof by contradiction can be used. Assuming that f is not injective, a set C and D can be constructed to violate the equality. This shows that f must be injective in order for the equality to hold.
  • #1
ELESSAR TELKONT
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0

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].
 
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  • #2


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


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

1. What does "intersection equality" mean in relation to a function being injective?

Intersection equality means that the intersection of the function's domain and its range is equal to the function's domain. In other words, every element in the function's domain must have a unique mapping to an element in the function's range.

2. How do you determine if a function is injective?

A function is injective if every element in its domain has a unique mapping to an element in its range. This can also be determined by checking if any two distinct elements in the domain have the same mapping in the range.

3. What is the significance of a function being injective?

A function being injective means that there are no "collisions" or duplicate mappings, making it easier to reverse the function and find the original input. This property is important in fields such as cryptography and data compression.

4. Can a function be injective if it is not one-to-one?

No, a function cannot be injective if it is not one-to-one. A function is one-to-one if every element in its domain has a unique mapping in its range, which is a requirement for a function to be injective.

5. How can you prove that a function is injective?

A common way to prove that a function is injective is to use the definition and show that every element in the domain has a unique mapping in the range. Additionally, you can also use a proof by contradiction where you assume that two distinct elements in the domain have the same mapping in the range, and then show that this leads to a contradiction.

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