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

  • #1
Eclair_de_XII
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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
Delta2
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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.
 
  • #3
Eclair_de_XII
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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."
 
  • #4
Delta2
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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.
 
  • #5
Delta2
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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.
 
  • #6
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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
Delta2
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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
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Let's say I proved this right. Would that make my counter-example wrong?
 
  • #9
Delta2
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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
Delta2
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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
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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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