5

I've got a little mistakes with that:

$A\subset f^{-1}(f(A))$ with equality if and only $f$ is injective.

For example, if we take $f(x)=x^2$ and $A=[-1,1]$, we have $$f(A)=f([-1,1])=\{f(x)\mid x\in[-1,1]\}=[0,1]$$ and $$f^{-1}(f(A))=f^{-1}([0,1])=\{x\mid f(x)\in[0,1]\}=[-1,1].$$

What's wrong here ?

Martin Sleziak
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idm
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  • See also http://math.stackexchange.com/questions/347880/proving-a-set-is-a-subset and http://math.stackexchange.com/questions/261157/show-s-f-1fs-for-all-subsets-s-iff-f-is-injective - you can find overview of some other similar results here. – Martin Sleziak Jan 18 '15 at 21:47
  • @idm Strange, if you interpret $f^{-1}$ as "pre-image" then the $\subset$ should be $\supset$. The question makes slightly more sense if we interpret $f^{-1}(B)$ as "forward image of $B$ under some (prospective) inverse function $f^{-1}$". In any case you need to provide more context to make the question clear. – Erick Wong May 27 '15 at 14:54

2 Answers2

5

I think you got mixed up. The statement is

$f\colon X → Y$ is injective if and only if for every subset $A ⊂ X$, $A = f^{-1}(f(A))$.

Not “for some subset”. The condition is, of course, always true for $A = X$.

k.stm
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5

The correct statement is with equality for all $A$ if and only $f$ is injective.

In your example if you pick $A=[0,1]$ then $f^{-1} (f(A)) \neq A$.

If $f$ is not injective, equality still holds for some $A$'s, but not for all.

Added If you actually mean that your proof is a proof of the statement, then here is what it is wrong with it.

You proved that this statement is true for the one function you chose and the one set you chose. But this doesn't prove that the statement is true for another function and another set. You need to prove it for ALL functions, not just one you chose.

N. S.
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  • ok, but what's wrong in my proof ? – idm Jan 18 '15 at 21:34
  • @idm Nothing is wrong. For some $A$'s you get $f^{-1}f(A)=A$, and you found one for which this happens...... Unless you think that your proof proves the statement, which is not true as a single example is not enough to prove a general statement.... – N. S. Jan 18 '15 at 21:36
  • ok, I see. Could you give me an example where we have not the equality ? – idm Jan 18 '15 at 21:37
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    @idm I already gave one, read my answer ;) – N. S. Jan 18 '15 at 21:39
  • I got it, tks :-) – idm Jan 18 '15 at 21:43