If the inverse image of the image of a set via some function is equal....

In summary, If the statement is "if there exists a ##X\subset A## such that ##f^{-1}(f(X))=X## then f is injective", then the validity of your counter example depends on the statement of the problem.
  • #1
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Homework Statement
to the set, is the function injective? Let ##f## be a function. Denote the domain ##A## and the target space ##B##. Let ##X\subset A##. If ##f^{-1}(f(X))=X##, then is ##f## injective?
Relevant Equations
##f(X)=\{f(x):x\in X\}##
##f^{-1}(f(X))=\{x\in A:f(x)\in f(X)\}##

Injective: If ##f## is ##\textbf{injective}##, then given any two points ##y_1,y_2\in f(X)##, the pre-images of these points must be equal.
##A=(-1,1)##
##B=[0,1)##
Define ##f:A\longrightarrow B## by ##f(x)=x^2##

Set ##X=A##.

##f(X)=\{f(x):x\in X\}=\{x^2:x\in(-1,1)\}=B##
##f^{-1}(f(X))=\{x\in A:f(x)\in f(X)\}=\{x\in (-1,1):x^2\in B\}=X##

Now choose a non-zero point ##y\in f(X)##. There are two pre-images of this point: ##x,-x\in (-1,1)##. This implies that ##f## is not injective even though the inverse image of the image of ##X## via ##f## is equal to ##X##.
 
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  • #2
It depends how exactly is the statement meant.

If it is meant for every ##X\subset A,f^{-1}(f(X))=X## then yes we can conclude from this that ##f## is injective.

However if the statement is meant as that exists one ##X\subset A## such that... then we can't conclude from this that ##f## is injective and that is what your counterexample proves.
 
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  • #3
Delta2 said:
If it is meant for every ##X\subset A,f^{-1}(f(X))=X## then yes we can conclude from this that ##f## is injective.
The statement to be proven can be formally stated as follows:

"Let ##f:A\longrightarrow B##. Then for all ##X\subset A##, if ##f^{-1}(f(X))=X##, then ##f## is injective."

I made an attempt to prove the negation using a counter-example.

Even more formally stated:

"For every function ##f## that maps a set ##A## to another set ##B##, for every subset ##X## contained in ##A##, if ##f^{-1}(f(X))=X##, then ##f## is injective."
 
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  • #4
This statement is true, it uses the for every (or for all) sentence which makes the difference.

Just consider as ##X=\{x\}## where x a random element of A. Then you can prove that for every ##x\in A## the preimage of ##f(x)## contains only one element, ##x##, therefore ##f## is injective.
 
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  • #5
To give you a better hint, you can prove that $$f(x_1)=f(x_2)\Rightarrow x_1=x_2$$ holds by considering the sets ##X_1=\{x_1\},X_2=\{x_2\}## and using that $$f(X_1)=f(X_2)\Rightarrow f^{-1}(f(X_1))=f^{-1}(f(X_2))$$ that is if two sets are equal, then their preimages are equal too.
 
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  • #6
Delta2 said:
This statement is true, it uses the for every (or for all) sentence which makes the difference.

Just consider as ##X=\{x\}## where x a random element of A. Then you can prove that for every ##x\in A## the preimage of ##f(x)## contains only one element, ##x##, therefore ##f## is injective.
I don't understand. You say that I must prove that the statement holds for all ##X\subset A##, yet you prove it by explaining that it holds for a specific subset of ##A##.
 
  • #7
Eclair_de_XII said:
I don't understand. You say that I must prove that the statement holds for all ##X\subset A##, yet you prove it by explaining that it holds for a specific subset of ##A##.
Nevermind my post #4 it is not well stated. Look at post #5. It is given that ##f^{-1}(f(X))=X## holds for every ##X\subset A##.
Since it holds for every subset X it holds for ##X_1=\{x_1\}## and ##X_2=\{x_2\}##. Then just proceed as post #5 says.
 
  • #8
Let's say I proved this right. Would that make my counter-example wrong?
 
  • #9
Eclair_de_XII said:
Let's say I proved this right. Would that make my counter-example wrong?
The validity of your counter example depends on the statement of the problem. If the statement was "if there exists a ##X\subset A## such that ##f^{-1}(f(X))=X## then f is injective"" then your counter example is right. But the statement is "if for all ##X\subset A##..." which makes your counter example wrong.
 
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  • #10
if we take ##f(x)=x^2## then the premise of the statement that is for all ##X\subset A,f^{-1}(f(X))=X## simply doesn't hold, it holds only for some X, not for every X.
 
  • #11
Thank you so much for the clarification. I very much appreciate you pointing out the oversight I had made in interpreting the statement I was meant to prove. I hope not to make such oversights again in proving theorems in the future.
 
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1. What does it mean for the inverse image of the image of a set to be equal?

When the inverse image of the image of a set is equal, it means that the function maps every element in the original set to a unique element in the image set, and the inverse function maps every element in the image set back to the original set.

2. How is the inverse image of the image of a set calculated?

The inverse image of the image of a set is calculated by first finding the image of the set using the given function, and then applying the inverse function to the resulting image set.

3. Can the inverse image of the image of a set ever be different from the original set?

No, the inverse image of the image of a set will always be equal to the original set as long as the function is one-to-one and onto. This means that every element in the original set has a unique mapping to an element in the image set, and every element in the image set has a corresponding mapping back to the original set.

4. How does the concept of inverse image of the image of a set relate to functions?

The inverse image of the image of a set is a concept that is specific to functions. It helps us understand the relationship between a function and its inverse, and how the two functions can be used to map elements between sets.

5. What are some real-life applications of the concept of inverse image of the image of a set?

The concept of inverse image of the image of a set has many applications in fields such as computer science, engineering, and physics. It is used in image and signal processing, data compression, cryptography, and many other areas where functions and their inverses are used to map and manipulate data.

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