one-one and onto proofs Prove that if $f$ takes the $[-1,1]$ onto $[-1,1]$, then 
$$f^{-1}(f(\{ 0 \})) = \{ 0 \}$$
Proof : Let the domain $= X$ which is the set $[-1,1]$ Let the co-domain $=Y$ which is the set $[-1,1]$.
$f$ onto implies that there exists an inverse function $g: Y \to X$ called the inverse S.T. $g(f(x))= x$ and $f(g(y))=y$. Since $0 \in X$ and $0 \in Y$ it follows directly that $g(f(0))=0$ and $f(g(y))=y$ which proves the claim.
Let $X,Y$ be sets and $f: X \to Y$. Prove that :
$$f(A\setminus B) = f(A)\setminus f(B)$$
TO be honest I am not even sure what this question is asking. It is very hard to imagine functions as cartesian products..
 A: 
f onto implies that there exists an inverse function

No.
Consider $f:\mathbb R \to [0, \infty)$ via  $f(x) = x^2$ or $g:\mathbb R \to [-1,1]$ via $f(x) = \sin x$.  These functions are onto but they are not one-to-one.  For any $f(x) = y$ there maybe two solutions to $x$ (if $x$ is a solution so is $-x$) and $\sqrt{x}$ is not an inverse function.  And if $g(x)  =y$ there will be infinitely many solutions (if $x$ is a solution, then so is $x + k2\pi$).  $\arcsin x$ is not an inverse function.
I realize mathematicians are not consistant with notation and that is their fault, not yours. but $f^{-1}(A)$ does not mean an inverse function but .... if $f:X\to Y$ and $A\subset Y$ then $f^{-1}(A)$ means all the elements of $X$ that get mapped do any element of $A$.
Example if $f(x) = x^2$ then $f^{-1}(\{16\}) = \{4,-4\}$ because $4$ and $-4$ are the elements so that $f(x) = 16$.  And $f^{-1}(\{16,25\}) = \{4,-4,5,-5\}$ because those are the elements that get mapped to $16$ or $25$.  
And $f^{-1}(\{-2\}) = \emptyset$ as nothing gets mapped to $-2$.  And $f^{-1}(\{-2, -25, 3, 49\}) = \{\sqrt 3,-\sqrt{3}, 7, -7\}$ because those are all the elements that get mapped to $-2,-25, 3,$ or $49$.
....
So your question.
$f$ being onto means there is $x\in [-1,1]$ so that $f(x) =0$.  There may be many of them.  There may be an infinite number of them.  But ther is one.  So $f^{-1}(\{0\})$ is not the empty set.
Now by definition $f^{-1}(\{0\}) = \{x\in [-1,1]| f(x) = 0\}$.
And be definition $f(A) = \{f(x)|x\in A\}$.
So.......
$f(f^{-1}(\{0\})) = f(\{x\in [-1,1]|f(x)=0\}) =$
$\{f(x)|x \in \{x\in[-1,1]|f(x)=0\}\}$ 
..... well if $x\in \{x\in[-1,1]|f(x)=0\}$ then that means $f(x) = 0$.
So $\{f(x)|x \in \{x\in[-1,1]|f(x)=0\}\}=$
$\{0\}$.
And that is it.
A: As to the second question.
If $X$ is a set then $f(X) = \{f(x)|x \in X\}$
So $f(A\setminus B) = \{f(x)| x \in A$ but $x\not \in B\}$.
And $f(A)\setminus f(B) = \{f(x)|x \in A\}\setminus \{f(x)|x\in B\}=$
Now the statement that $f(A\setminus B) = f(A)\setminus f(B)$ is not true.  It is true if $f$ is one-to-one but it is not true in general.
If $w \in f(A\setminus B)$ then there is an $x\in A\setminus B$, that is that $x \in A$ but $x \not \in B$ so that $f(x) = w$.
So $w \in f(A)$ because $w = f(x)$.  Now $x \not \in B$ but there may or may not be a $y\in B$ so that $f(y) =w$ as well.  If there is, then $w \in F(B)$.  If there isn't then $w\not \in f(B)$.  If $f$ is one-to -on then there isn't any but if $f$ isn't one-to-one there might be.
So it might be that $w \in f(A)\setminus f(B)$ or it might be $w \not \in f(A)\setminus f(B)$.  If $w \not \in f(A)\setminus f(B)$ then $f(A\setminus B)\ne f(A)\setminus f(B)$.
But if $f$ is one-to-one then $w$ can't be in $f(B)$ because $x\in A\setminus B$ is the only solution to $f(x)=w$.  This is true for all $w\in f(A\setminus B)$ so $f(A\setminus B)\subset f(A)\setminus f(B)$.  IF $f$ is one-to-one.
On the other hand if $w \in f(A)\setminus f(B)$ then $w \in F(A)$ and $w \not \in F(B)$.  So there is an $x \in A$ so that $w = f(x)$.  And there are absolutely no $y\in B$ so that $f(y) =w$.  So the $x\in A$ can not be in $B$.  So $x \in A\setminus B$.  And $f(x) =w$ so $w\in f(A\setminus B)$.  That is true for all $w \in f(A)\setminus f(B)$ so $f(A)\setminus f(B)\subset f(A\setminus B)$.  
This is true of all functions.
So if $f$ is one-to-one this is true as mutual subsets are equal.  But if $f$ is not one to one it might not be true as its possible for $f(x) =f(y) = w$ where $x \in A\setminus B$ and $y \in B$.
