I want to generalize the following result: if $(X , d)$ is a metric space, $f : (X , d) \to (X , d)$ is an homeomorphism and $A \subset X$, then $f(X \setminus A) = X \setminus f(A)$, so I have thought that if $X$ is an abstract set, $f : X \to X$ is a suprajective map and $A \subset X$, then also $f(X \setminus A) = X \setminus f(A)$ and I have tried to prove that: we begin proving $f(X \setminus A) \subset X \setminus f(A)$, so let $y \in f(X \setminus A)$. On the one hand, by the suprajectivity of $f$, exists $x \in X \setminus A$ such that $f(x) = y$, so $f(x) \notin f(A)$ because $x \notin A$. Equivalently, $y \in X \setminus f(A)$. To prove $X \setminus f(A) \subset f(X \setminus A)$, let $z \in X \setminus f(A)$. Again, by the suprajectivity of $f$, exists $w \in X$ such that $f(w) = z \notin f(A)$, so $w \notin A$ and it proves that $z \in f(X \setminus A)$; in fact, if $w \in A$, then $f(w) \in f(A)$, so $z \in f(A)$. Do you think it is correct? Thank you very much.
$f(X \setminus A) = X \setminus f(A)$?
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0What is suprajective map? – 2017-02-20
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0$\varphi : Y \to Z$ is a suprajective map if, given $z \in Z$, exists $y_z \in Y$ such that $\varphi(y_z) = z$, and it happens for all $z \in Z$. – 2017-02-20
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0What you could use is that $f$ is bijective (and has a bijective inverse $f^{-1}$) and use that to show that $f(X\setminus A)$ and $X\setminus f(A)$ has the same members. – 2017-02-20
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0Okay good to know. I was taught that it is called as Surjective map. :) – 2017-02-20
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0@skyking: if my proof is fine, I would like take my proof because I want generalize as much as possible this result and a surjective map is more general than a bijective map. If there is an error in my proof, I will try to make you say; thank you. – 2017-02-20
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0Actually you didn't use suprajectivity in first part. If $y \in f(X \setminus A)$ then there has to be an element $x \in X \setminus A$ such that $f(x)=y$. – 2017-02-20
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0And what can I use to prove that? – 2017-02-20
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0What you used was that $y \in \text{Range}f$. – 2017-02-20
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0I think we don't know if $\mbox{Range } f$ exists because $X$ is an abstract set, not necessarily a vector space. – 2017-02-20
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0$y \in f(X \setminus A)$ means $y \in \{f(z) : z \in X \setminus A \}$. Now clear? – 2017-02-20
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0Yes, it is clear, so you say that not necessarily exists $x \in X \setminus A$ such that $f(x) = y$, right? – 2017-02-20
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0No no no! That I didn't imply. I just tried to make your reasoning of that part more logical. Even if $f$ was not given to be suprajective, that part which you proved holds true! – 2017-02-20
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0I fully understand what you mean. Since my reasoning is not perfect, can you help me reason why exists $x \in X \setminus A$ such that $f(x) = y$, please? I haven't understood very well that it happens because $y \in \{f(z) : z \in X \setminus A\}$. – 2017-02-20
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0@skyking: I think quite stupid is don't know to read fine: I said "if my proof is fine,..." – 2017-02-20
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0@joseabp91 Well, yes, I misread your answer. Then it's of course not stupid. I appologize. Your proof is not fine however (see my answer for details). – 2017-02-20
1 Answers
No your proof is not correct. You actually need to use the fact that $f$ is bijective (ie both surjective and injective), surjective is not enough.
Take for example $X = \mathbb R$ and $A=[-1,1]$ and $f(x) = x\sin(\pi x/2)$, the function is certainly surjective. Now $f(X\setminus A) = X \ne X \setminus f(A)$. This function however is not bijective.
The actual error happens when you try to prove that $f(X\setminus A)\subset X\setminus f(A)$. It's the actual part where injectivity is required. The exact error is that you conclude that $f(x)\notin f(A)$ because $x\notin A$. On the other hand surjectivity is not required to prove this part.
To prove it you observe that since $y\in f(X\setminus A)$ there exists an $x\in X\setminus A$ such that $f(x) = y$ (by definition of the image of a set). Now to conclude that $f(x)\notin f(A)$ we use the fact that if it were in $f(A)$ we would have a $\xi\in A$ such that $f(\xi)=f(x)$ (by definition of image of a set), but due to $f$ being injective that would mean that $x = \xi$ since $f(x)=f(\xi)$ which would mean that $x\in A$ which contradicts $x\in X\setminus A$ - so we can conclude that $f(x)\notin f(A)$.
To sum up with $f$ being surjective we can conclude that $f(X\setminus A)\supset X\setminus f(A)$ and if $f$ is injective we can conclude that $f(X\setminus A)\subset X\setminus f(A)$.
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0Ok I understand thank you. So you think I must use also surjectivity? Then I have to suppose that $f$ is bijective? – 2017-02-20
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0Do you think so that my proof of $f(X \setminus A) \subset X \setminus f(A)$ is correct? – 2017-02-20
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0Thanks for showing my error. Now I understood why the condition bijective is necessary. @joseabp91 Apologies for guiding you on a wrong path where I mistook that only suprajectivity is sufficient to prove the entire thing. As this answer tells, you need both injective and surjective! – 2017-02-21
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0@joseabp91 Yes, maybe it's a bit out of order: "in fact, if $w \in A$, then $f(w) \in f(A)$, so $z \in f(A)$" is an argument for the conclusion "so $w \notin A$" - it would be more obvious if it was noted closer to that. – 2017-02-21
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0I think you want to say in the last sentence if $f$ is injective, then $f(X \setminus A) \subset X \setminus f(A)$ and if $f$ is surjective, then $f(X \setminus A) \supset X \setminus f(A)$, right? – 2017-02-21
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0@joseabp91 Yes, that's correct - I've modified the last sentence... – 2017-02-21
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0So are you sure that we need surjectivity of $f$? – 2017-02-21
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0@joseabp91 Yes, assume that $A=\emptyset$, then the statement becomes $f(X)\supset X$ which only can be true if $f$ is surjective. – 2017-02-21
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0Ok thank you very much – 2017-02-21