$f:A\to B$ is given, $C_1$ and $C_2$ are subsets of $A$. Is it true that $f(C_1 \cap C_2) = f(C_1) \cap f(C_2)$?
I can give an example to prove it wrong: if we take $C_1=[-1,0]\,,\, C_2=[0,1]$ , and if $f(x)=x^2$ , then, I know that $[0,1]≠\{0\}$ therefore the statement is wrong. However, I tried to prove it and:
From left to right:
$x\in f(C_1 \cap C_2) \Rightarrow f^{-1}(x) \in C_1 \cap C_2 \Rightarrow f^{-1}(x) \in C_1\;\; \text{and}\;\; f^{-1}(x) \in C_2 \Rightarrow $
$ x\in f(C_1) \;\;\text{and}\;\; x\in f(C_2) \Rightarrow x\in f(C_1)\cap f(C_2)$
Therefore it is true. Let's check from right to left:
$x∈f(C_1)\cap f(C_2) \Rightarrow x∈f(C1)\;\;\text{ and}\;\; x∈f(C_2) \Rightarrow f^{-1}(x)∈C_1\;\; \text{and}\;\; f^{-1}(x)∈C_2$
$ \Rightarrow f^{-1}(x)∈C_1\cap C_2 \Rightarrow x∈f(C_1\cap C_2)$
This seems true also. However it should not be. What is wrong can you see?