# Finding a function

#### onemore

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

#### kompik

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

#### novice

$$\displaystyle f(f^{-1}(D))$$ 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.

#### Swlabr

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 $$\displaystyle D = \{6\}$$.

novice

#### novice

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

MHF Hall of Honor
novice said:
kompik said:
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.

novice

#### novice

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?

#### novice

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.

#### novice

$$\displaystyle \frac{a}{b} \mapsto ab$$ 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)

#### Swlabr

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. $$\displaystyle f^{-1}(f(D)) = f^{-1}(f(6)) = f^{-1}(6) = \{6, 2/3, 3/2\} \neq D$$.

novice