[tex]A_0 \subset f^{-1} (f (A_0)) [/tex] 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?
Ok I think I got it. If we don't know that [tex]f:A \rightarrow B[/tex] is bijective or even surjective/injective, we want [tex]f^{-1}[/tex] to be [tex] \{ a | f(a) \in B\}[/tex] is this correct? Let [tex]f:A \rightarrow B[/tex] and [tex] A_0 \subset A [/tex] Say we want to show that [tex]A_0 \subset f^{-1}( f(A_0)) [/tex] Suppose we have [tex]a \in A_0 [/tex] then by the definition of a function [tex] f(a) = b [/tex] for some [tex]b \in B[/tex] [tex]f^{-1}(b) [/tex] then is [tex]\{ c | f(c) =b\}[/tex] since we have already established that [tex] f(a) = b [/tex] it is clearly the case that [tex] a \in \{ c | f(c) =b\} = f^{-1}(f(a))[/tex]. Therefore, since we choose [tex]a[/tex] arbitraraly [tex] A_0 \subset f^{-1}(f(A_0))[/tex] Is this right?
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.}
Wow, that's discouraging. Anyways, I think I said your exact definition of [tex]f^{-1}[/tex] in my second post. Where I said if [tex]f:A \rightarrow B[/tex] "we want [tex]f^{-1}[/tex] to be [tex]\{a | f(a) \in B \} [/tex]" How was my proof of [tex] A_0 \subset f^{-1} (f(A_0))[/tex]? Was that any good? If not I hope it was at least, yet again, humorous...