So, I can't figure out if the following statement is true or not:
For a function $f:X \rightarrow Y$, if $f$ is one-to-one, then $f(A^C)=(f(A))^C$ where $B \subset X$.
I have an example below which suggests that this is not always true. However, I've recently had several people (without proof) tell me that it is true. So I honestly don't know. Observe my example below:
Define $f:X\rightarrow Y$ where $X=\lbrace 1, 2 \rbrace$, $Y=\lbrace 1, 2, 3 \rbrace$, in such a way that $f(\lbrace 1 \rbrace) = \lbrace 1 \rbrace$ and $f(\lbrace 2 \rbrace) = \lbrace 2 \rbrace$. Let $A\subset X$. Now let $A=\lbrace 1 \rbrace\rightarrow f(A)=\lbrace 1 \rbrace$. So that means that $A^C=\lbrace 2 \rbrace \rightarrow f(A^C)= \lbrace 2 \rbrace$. Yet $(f(A))^C= \lbrace 2, 3 \rbrace$.
This seems to disprove the statement… right? Or is there something fundamental about complementation that I'm not understanding here?
Best Answer
You're right. For injective functions, the image of the complement is only guaranteed to be a subset of the complement of the image, and the example you've given proves that.
If $f$ is surjective as well as injective, however, then the image of the complement and the complement of the image are the same (this is a good exercise).