Help with set bijection proof Let $f: M \to N$ be a bijection
Let $A, B \subset M$
Then show:


*

*$f(A \cap B) = f(A)\cap f(B)$

*$f(A^c) = f(A)^c$
I have no idea how to do this honestly!
How do you go about doing this? Can you draw a picture to solve these types of problems? Must $f$ be a bijection?
 A: The typical way to show that two sets $U$ and $V$ are equal is to show that $U \subset V$ and $V \subset U$.
For your first equality, let $y \in f(A \cap B)$.  Since $f$ is a bijection, $f^{-1} : N \to M$ is well-defined, and there is a unique $x = f^{-1}(y)$.  By assumption, $x \in A \cap B$. Then since $x \in A$, $y \in f(A)$, and since $x \in B$, $y \in f(B)$.  Thus $y \in f(A) \cap f(B)$, and so $f(A \cap B) \subset f(A) \cap f(B)$.
Now let $y \in f(A) \cap f(B)$.  Since $y \in f(A)$, $x \in A$, and since $y \in f(B)$, $x \in B$.  Thus $x \in A \cap B$, so $y = f(x) \in f(A \cap B)$, and so $f(A) \cap f(B) \subset f(A \cap B)$.
You should check the second equality similarly.
A: *

*($\subseteq$ direction):
Let $ y \in f(A\cap B)$. Then there exists $x \in A\cap B$ such that $y = f(x)$. Since $x \in A$, it follows that $y \in f(A)$ and since $x \in B$, it follows that $y \in f(B)$. Since $y$ was arbitrary, this shows that $f(A\cap B) \subseteq f(A)\cap f(B)$.


($\supseteq$ direction): Let $y \in f(A) \cap f(B)$. Then there exists $x_1 \in A$ and $x_2 \in B$ with $f(x_1) = f(x_2) = y$. But $f$ is injective since it is a bijection. It follows that $x:= x_1 = x_2$, $x \in A \cap B$, and $f(x) = y$ so that $y \in f(A\cap B)$. Since $y$ was arbitrary, this shows that $f(A) \cap f(B) \subseteq f(A\cap B)$. $\square$ 


*($\subseteq$ direction): Let $y \in f(A^c) $. Then there exists $x \in A^c$ with $f(x) = y$. Since $f$ is injective, $f$ does not map any other points to $y$ besides $x$. It follows that $y \not \in f(A)$. Hence $y \in f(A)^c$. Since $y$ was arbitrary, this shows that $f(A^c) \subseteq f(A)^c$.


($\supseteq$ direction): Let $y \in f(A)^c$. Then for all $x$ in $A$, $y \not = f(x)$. However, $f$ is surjective since it is a bijection, so there is some $x$ in $M$ with $f(x) = y$. This $x$ must be not be in $A$, so $x \in A^c$. Hence $y \in f(A^c)$. Since $y$ was arbitrary, this shows that $f(A)^c \subseteq f(A^c)$. $\square$
