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 throught 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.