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

[Eclair_de_XII]

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 $X$.

 

[Delta2]

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.

 

[Eclair_de_XII]

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."

 

[Delta2]

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.

 

[Delta2]

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.

 

[Eclair_de_XII]

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 $A$.

 

[Delta2]

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.

 

[Eclair_de_XII]

Let's say I proved this right. Would that make my counter-example wrong?

 

[Delta2]

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.

 

[Delta2]

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.

 

[Eclair_de_XII]

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.