Is is true $\forall_{A,B \subset X}: f(A - B)=f(A) - f(B)$? In C. Adam's Topology, it is written that $f(A) - f(B) \subset f(A - B)$ for any function, but $f(A) - f(B) = f(A - B)$ iff f is bijective. I can come the half way, i.e., $f(A) - f(B) \subset f(A - B)$; but for the $f(A - B) \subset f(A) - f(B)$, I don't know both of how to prove correctness of it for bijective and falseness of it for non-one-to-one function;
Suppose $y\in f(A) - f(B)$, then $y$ belongs to $f(A)$ but it does not belong to $f(B)$; thus there exist(s) one or more $x_i$ such that $y=f(x_i)$ and non of $x_i$ must be in $B$ since it results in $f(x_i)=y\in f(B)$, a contradiction; so for arbitrary $y$, if $y\in f(A) - f(B) \implies y\in f(A - B)$, thus $f(A) - f(B) \subset f(A - B)$.
I appreciate it if please help me to prove(one-to-one)/disprove(non-one-to-one) in case of $f(A - B) \subset f(A) - f(B)$.       
 A: For the bijective case, consider :

$x \in f(A - B)$;

this means that there is $y_1 \in A - B$ such that : $x=f(y_1)$, and $y_1 \in A - B$ implies : $y_1 \in A$ and $y_1 \notin B$.
From : $x=f(y_1)$ for $y_1 \in A$ we have that : $x \in f(A)$.
Assume now that : $x \in f(B)$; this implies that there exists $y_2 \in B$ such that : $x=f(y_2)$.
If $y_2 = y_1$, we have that : $y_1 = y_2 \in B$, contradicting the above fact that : $y_1 \notin B$.
Thus we have $y_2 \ne y_1$ and :

$f(y_1)= x = f(y_2)$.

But according to the definition of bijection :

for a pairing between $X$ and $Y$ to be a bijection, four properties must hold:
[...] iv) no element of $Y$ may be paired with more than one element of $X$.

Thus we have a contradiction, because $x$ is paired with two distict elements $y_1$ and $y_2$.
Conclusion : we have to discard the assumption that : $x \in f(B)$, and thus : $x \notin f(B)$.
From : $x \in f(A)$ and $x \notin f(B)$ we conclude with : $x \in f(A) - f(B)$, and thus we have :


$f(A - B) \subset f(A) - f(B)$.


A: For a counterexample when $f$ is not bijective : Let $f:\mathbb R \to \mathbb R : x \mapsto 1$. Take $A = \mathbb R$ and $B = \{0\}$. Then $f(A \setminus B) = \{1\}$ but $f(A) \setminus f(B) = \{1\} \setminus \{1\} = \emptyset$.
A: Let $y\in f(A-B)$ and note that $f(A-B)\subseteq f(A)$ as a consequence of $A-B\subseteq A$. So automatically we have $y\in f(A)$. Some $a\in A-B$ must exist with $y=f(a)$. Now suppose that also $y=f(b)$ for some $b\in B$. Then injectivity of $f$ provides a contradiction and we can conclude that it implies that $y\notin f(B)$. That leads to $y\in f(A)-f(B)$. 
Proved is now that $f(A-B)\subseteq f(A)-f(B)$ if $f$ is injective. Actually for this $f$ does not have to be a bijection (injectivity is enough).
Conversely let it be that for every pair $A,B$ of subsets of $X$ we have $f(A-B)\subseteq f(A)-f(B)$ and let $u,v\in X$ such that $f(u)=f(v)$. Define $A=\{u\}$ and $B=\{v\}$. Then $f(A-B)=f(A)-f(B)=\emptyset$ implying that $A-B=\emptyset$ or equivalently $u=v$. Proved is now that under the conditions mentioned $f$ is injective.
