How to prove that: $f(A\cap B) \subseteq f(A) \cap f(B)$, and some few others With $f:X \rightarrow Y; A, B \subseteq X; C, D \subseteq Y$, I'm given these identities:
1. $f(A\cap B) \subseteq f(A) \cap f(B)$
2. $f(A\setminus B) \subseteq f(A) \setminus f(B)$
3. $f^{-1}(C\cap D) \subseteq f^{-1}(C) \cap f^{-1}(D)$
4. $A \subseteq f^{-1}(f(A))$
5.$f^{-1}(f^{-1}(C)) = C \cap f(X) \subseteq C$
Here's what I need to do:
a)Find examples of functions and non-empty sets A, B for which 1. and/or 2. aren't equal (i.e. they are $\subset$, but not $=$)
I believe this is the case for any non-injective function for which $a \neq b$ but $f(a) = f(b)$ , like $f(x) = |x|$, with $a = 1$ and $b=-1$. I'd like to hear some more examples of functions like this.
b)Here I need to find the same thing like a) but for identities 4 and 5, however I'm not sure what even to look for.
c) Prove 1 or 2 and 3 or 4. Here's my not so successful attempt at proving 1:
$f(A \cap B) = \{y \in Y : y = f(x) \wedge x \in A \cap B\} = \{y \in Y : y = f(x) \wedge (x \in A \wedge x \in B)\} = \{y \in Y : (y = f(x) \wedge x \in A) \wedge (y = f(x) \wedge x \in B)\} = \{y \in Y : y = f(x) \wedge x \in A\} \cap \{y \in Y : y = f(x) \wedge x \in B\} = f(A) \cap f(B)$
And I made an analogous "proof" for 2, however, obviously it's not right as I get that $f(A \cap B) = f(A) \cap f(B)$ instead of $\subseteq$. And for the second part of c) (i.e. 3 and 4), I'm not even sure how to start.
Any help at solving a) b) and c) is highly appreciated.
 A: 1): $A\cap B\subseteq A\Rightarrow f\left(A\cap B\right)\subseteq f\left(A\right)$
and $A\cap B\subseteq B\Rightarrow f\left(A\cap B\right)\subseteq f\left(B\right)$
. Consequence: $$f\left(A\cap B\right)\subseteq f\left(A\right)\cap f\left(B\right)$$
2) is not true: let $f$ be a constant function and let
$A\backslash B\neq\emptyset\wedge B\neq\emptyset$. Then $f\left(A\backslash B\right)\neq\emptyset$
and $f\left(A\right)\backslash f\left(B\right)=\emptyset$
3): $$x\in f^{-1}\left(C\cap D\right)\iff f\left(x\right)\in C\cap D\iff$$$$ f\left(x\right)\in C\wedge f\left(x\right)\in D\iff x\in f^{-1}\left(C\right)\cap f^{-1}\left(D\right)$$
4): $$x\in A\Rightarrow f\left(x\right)\in f\left(A\right)$$
combined with: $$f\left(x\right)\in f\left(A\right)\iff x\in f^{-1}\left(f\left(A\right)\right)$$
Adjusted 5): To be proved is probably $f\left(f^{-1}\left(C\right)\right)=C\cap f\left(X\right)$
(the notation $f^{-1}\left(f^{-1}\left(C\right)\right)$ makes no
sense). 
$$y\in f\left(f^{-1}\left(C\right)\right)\iff y=f\left(x\right)\text{ for some }x\in f^{-1}\left(C\right)\iff y\in C\cap f\left(X\right)$$
A: for a), consider the function $f(x) = x^2 $ and put $A = [-1,0) $ and $B = (0,1] $, then 
$$ A \cap B = \varnothing \implies f( A \cap B ) = \varnothing $$
but, $f(A) = f(B) = (0,1] $, hence
$$ f(A) \cap f(B) = (0,1] $$
A: To show $X\subseteq Y$, it is often easiest to show that $x\in X\implies x\in Y$. For example, if $x\in f(A\cap B)$, then $x=f(y)$ for some $y\in A\cap B$. Then $y \in A$, so $x\in f(A)$, and similarly $x\in f(B)$. Hence, $x\in f(A)\cap f(B)$. 
