Question Regarding Sets and Functions

[Diffy]

$$A_0 \subset f^{-1} (f (A_0))$$

This inclusion is an equality if f is injective.

What I can't understand is how it is even defined if f isn't a bijection. If it is not a bijection, then there is no inverse function. Is there?

 

[Diffy]

Ok I think I got it. If we don't know that $$f:A \rightarrow B$$ is bijective or even surjective/injective, we want $$f^{-1}$$ to be $$\{ a | f(a) \in B\}$$

is this correct?

Let $$f:A \rightarrow B$$ and $$A_0 \subset A$$

Say we want to show that $$A_0 \subset f^{-1}( f(A_0))$$

Suppose we have
$$a \in A_0$$
then by the definition of a function $$f(a) = b$$ for some $$b \in B$$
$$f^{-1}(b)$$ then is $$\{ c | f(c) =b\}$$ since we have already established that $$f(a) = b$$ it is clearly the case that $$a \in \{ c | f(c) =b\} = f^{-1}(f(a))$$. Therefore, since we choose $$a$$ arbitraraly $$A_0 \subset f^{-1}(f(A_0))$$

Is this right?

 

[HallsofIvy]

Okay, I won't laugh at you too hard!

The very first time I had to present a proof before the class in a graduate class it was something exactly like this! I went through the whole thing, assured that I was exactly right! I did the whole proof assuming that f HAD an inverse! Very embarrasing! It's probably the one thing I remember more than anything else from my graduate student days!

f^-1(A), where A is a set, is defined as {x| f(x) is in A}. No, it is not required that f be "one-to-one"! If, for example, f(x)= x^{2, where f is surely not one-to-one, then f-1([-1,4]= {all x such that f(x) is in that set}. That, of course is the interval [-2, 2] since f(-2)= f(2)= 4 and all numbers between -2 and 2 are taken to numbers between 0 and 4 and so between -1 and 4.}

 

[Diffy]

Okay, I won't laugh at you too hard!

Wow, that's discouraging.

Anyways, I think I said your exact definition of $$f^{-1}$$ in my second post. Where I said if $$f:A \rightarrow B$$ "we want $$f^{-1}$$ to be $$\{a | f(a) \in B \}$$"

How was my proof of $$A_0 \subset f^{-1} (f(A_0))$$? Was that any good? If not I hope it was at least, yet again, humorous...

 

[Eighty]

Both your definition and proof are correct.