Disprove the statement $f(A \cap B) = f(A) \cap f(B)$ If someone could walk me through this I would greatly appreciate it.
Disprove the following statement: 
If $f : X \rightarrow Y$ is a function and $A$, $B$
are subsets of $X$ then $f(A \cap B) = f(A) \cap f(B)$.
 A: The other answers find counterexample for $f(A\cap B) =f(A)\cap f(B)$, but it is easy to show that $f(A\cap B)\subseteq f(A)\cap f(B)$:
If $A\subseteq B$, then $f(A)\subseteq f(B)$. 
Now, $A\cap B\subseteq A, B$, and thus $f(A\cap B)\subseteq f(A), f(B)\implies f(A\cap B)\subseteq f(A)\cap f(B)$.
Actually, equality holds if and only if $f$ is injective.
Let $f$ be injective and take $y\in f(A)\cap f(B)$. Then, by the definition of image there are $x_A\in A$ and $x_B\in B$ such that $y = f(x_A) = f(x_B)$. By injectivity $x_A = x_B\in A\cap B$. But then $y\in f(A\cap B)$. Hence, $f(A)\cap f(B) \subseteq f(A\cap B)$ and then by the above $f(A)\cap f(B) = f(A\cap B)$.
If $f$ is not injective, choose $x\neq y$ such that $f(x) = f(y)$. Then $f(\{x\}\cap\{y\}) \neq f(\{x\})\cap f(\{y\})$.
A: $$A=\{0,-1\} \,\, \, , \,\, \, B=\{0,1,-2\}\,\,, \,\, C=\{0,1,-1,-2\}\,\,,D=\{1,2,3\}$$ 
let 
$$f:C\to D$$
such that
$$f=\{(0,1),(1,1),(-1,3),(-2,3)\}$$
we have
$f(A)=\{1,3\}$ , $f(B)=\{1,3\}$ , $f(A\cap B)=\{1\}$ whereas $f(A)\cap f(B)=\{1,3\}$
Generally we have
$$f(A\cap B)\subseteq f(A)\cap f(B)$$
A: Just to give a "minimalist" counterexample that clearly makes use of the hints in the comments: consider $X = \{0,1\}$, let $A = \{0\}$ and $B = \{1\}$. Take $f: X \to X: x \mapsto 0$. Then
$$f(A\cap B) = f(\emptyset) = \emptyset$$
But 
$$f(A) \cap f(B) = \{0\} \cap \{0\} = \{0\}$$
A: $f(x) = x^2$, $A = [-1,0],B = [0,1]$.
$f(A\cap B) = f(\{0\}) = \{0\}$
But $f(A)\cap f(B) = [0,1]$
