Finding a function

Mar 2010
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Find an example of a function \(\displaystyle f: A \rightarrow B \) and a subset \(\displaystyle D \subset B \) such that \(\displaystyle f(f^{-1} (D)) \neq D \).


Im totally lost on this one. Any help please
 
Mar 2010
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Bratislava
Find an example of a function \(\displaystyle f: A \rightarrow B \) and a subset \(\displaystyle D \subset B \) such that \(\displaystyle f(f^{-1} (D)) \neq D \).


Im totally lost on this one. Any help please
What about this example: Choose any function that is not surjective for f and put D=B.
 
Sep 2009
502
39
What about this example: Choose any function that is not surjective for f and put D=B.
\(\displaystyle f(f^{-1}(D))\) means that \(\displaystyle f\) is bijective. So we cannot use \(\displaystyle f\) that is non-surjective. Besides, if we want to create a non-surjective function, the best is to make the range such that \(\displaystyle |D|<|B|, \) where \(\displaystyle D\subset B\). It will automatically be non-surjective.
 
May 2009
1,176
412
Find an example of a function \(\displaystyle f: A \rightarrow B \) and a subset \(\displaystyle D \subset B \) such that \(\displaystyle f(f^{-1} (D)) \neq D \).


Im totally lost on this one. Any help please
What do you mean by \(\displaystyle f^{-1}\)? Is it that \(\displaystyle f^{-1}(S) = \{s : f(s) \in S\}\)? As clearly this result does not hold if this is your definition of \(\displaystyle f^{-1}\), however this is the `natural' definition...

(The result does not hold as \(\displaystyle f^{-1}(D) = \{a:f(a) \in D\}\) so \(\displaystyle f(f^{-1}(D)) = f(f^{-1}(\{a:f(a) \in D\})) = f(a) \in D\).)

Now, \(\displaystyle f^{-1}(f(D)) \neq D\) does hold sometimes. Take, for example, \(\displaystyle f: \mathbb{Q} \rightarrow \mathbb{Q}\), \(\displaystyle \frac{a}{b} \mapsto ab\) and \(\displaystyle D = \{6\}\).
 
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Sep 2009
502
39
I don't quite understand the following:
Now, \(\displaystyle f^{-1}(f(D)) \neq D\) does hold sometimes. Take, for example, \(\displaystyle f: \mathbb{Q} \rightarrow \mathbb{Q}\), \(\displaystyle \frac{a}{b} \mapsto ab\) and \(\displaystyle D = \{6\}\).
I can't picture it. Could you please elaborate a little?
 

emakarov

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Oct 2009
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novice said:
kompik said:
What about this example: Choose any function that is not surjective for f and put D=B.
means that is bijective. So we cannot use that is non-surjective.
I agree with the idea of kompik. On the other hand, \(\displaystyle f(f^{-1}(D))\) is not a proposition, so it does not mean anything. In any case, I don't see where it follows that \(\displaystyle f\) must be bijective.
 
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Sep 2009
502
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I agree with the idea of kompik. On the other hand, \(\displaystyle f(f^{-1}(D))\) is not a proposition, so it does not mean anything. In any case, I don't see where it follows that \(\displaystyle f\) must be bijective.
Could you paint me a picture? A function that have an inverse function, does it not return the preimages?
 
Sep 2009
502
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I agree with the idea of kompik. On the other hand, \(\displaystyle f(f^{-1}(D))\) is not a proposition, so it does not mean anything. In any case, I don't see where it follows that \(\displaystyle f\) must be bijective.
I see you point now.
 
Sep 2009
502
39
\(\displaystyle \frac{a}{b} \mapsto ab\) and \(\displaystyle D = \{6\}\).
Quite clever. I see a function taking a point from a plane and put it on a number line, while the inverse cannot cannot take the same point from the number line and put it back onto the plane.(Evilgrin)
 
May 2009
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Quite clever. I see a function taking a point from a plane and put it on a number line, while the inverse cannot cannot take the same point from the number line and put it back onto the plane.(Evilgrin)
Not quite...the way we defined "inverse" is not a single element, it will give a set. \(\displaystyle f^{-1}(f(D)) = f^{-1}(f(6)) = f^{-1}(6) = \{6, 2/3, 3/2\} \neq D\).
 
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