We have the following question for our homework-
Determine whether each of the following statements is true or false. If the statement is true, prove your answer. (Admittedly, some of these are hard to provide a proof for that is anything more than just citing a definition, but do your best.) If the statement is false, provide a counterexample. (In the statements below, $f:A\rightarrow B$, $E\subseteq A$ and $S\subseteq B$.)
a If $x\in f^{-1}(S)$, then $f(x)\in S$.
b If $y\in S$, then there exists $x\in f^{-1}(S)$ such that $f(x) = y$.
c If $y\in f(E)$, then there exists $x\in E$ such that $f(x) = y$.
d If $y\in f(f^{-1}(S))$, then $y\in S$.
e If $x\in f^{-1}(f(E))$, then $x\in E$.
I have an answer for each, but I'm not sure if this is correct (some of my friends got different answers).
a. False. Let $f^{-1}(x)= x^2$. Thus $f(x)= \sqrt{x}$. For $x=4 $, we have both $f(x)=2$ and $f(x)=-2$, but 2 and -2 can be in different sets.
b. True. This is the definition of an inverse.
(c)~True. Since y is in f(E), every value,x, in E must have a corresponding match such that $f(x)=y$
d. False. Is only true for surjective functions, since we $f^{-1}(S)$ might not exist always.
e. False. Only true for injective functions.
Am I correct? If not, where am I wrong? Any input would help.
a) True, actually, by the definition of a pre-image. Your counterexample doesn't work because $f(x) =x^2$ is not invertible, so $f^{-1}(x)$ doesn't exist (at least if you're defining it as $f: \mathbb{R} \rightarrow \mathbb{R}$).
b) False, because $S$ may not be mapped to by any element
c) True
d) True. As Randall mentioned, the pre-image of a set $f^{-1}(S)$ always exists even though $f^{-1}(x)$ may not. $y$ is in the image of the pre-image of $S$, and since functions only map each $x$ to one $f(x)$, it must be that $y \in S$.
e) False. If we take the image of a set $S$, and then take its pre-image, we will end up with $S$ and other elements that mapped to the same things $S$ did. So $f^{-1}(f(E))$ contains more elements other than $E$.