Can you please check my proof of the following theorem?
Theorem: Let $f:A \rightarrow B$ be a bijective map, and let $P,Q \subseteq A$ be any sets. Then $f(P-Q)=f(P)-f(Q)$.
Proof: Let $x \in f(P)-f(Q)$. Hence $x \in f(P)$ and $x \notin f(Q)$. From the former, we deduce that there exists some $a \in P$ such that $f(a)=x$. Let $a_0 \in P$ that element. We know that $f(a_0) \notin f(Q)$, so $a_0 \notin Q$. Then we have that $a_0 \in P-Q$ and it follows that $x=f(a_0) \in f(P-Q)$. Therefore $f(P)-f(Q) \subseteq f(P-Q)$.
Let $y \in f(P-Q)$. Since $f$ is bijective, there is only one $b \in P-Q$ such that $f(b)=y$. Let $b_0 \in P-Q$ be that element. We know that $b_0 \in P-Q$, so $b_0 \in P$ and $b_0 \notin Q$. From that we conclude that $f(b_0) \in f(P)$ and $f(b_0) \notin f(Q)$. Hence $f(b_0) \in f(P)-f(Q)$, so $y \in f(P)-f(Q)$. Therefore $f(P-Q) \subseteq f(P)-f(Q)$. $\square$
I'm not sure about the second part of the proof. My doubts are:
I'm not quite sure if we can deduce that $f(b_0) \notin f(Q)$ from $b_0 \notin Q$.
The fact that $f$ is bijective allows us to conclude that such $b_0$ is unique, and (for example) it could never be the case of having one element in $P$ and other in $Q$ with the same image (what would implied that such image could be in $f(P-Q)$ but not in $f(P)-f(Q)$), right?
Thank you for your attention!
using the definition of image you have $a \in f(Q)$ if and only if there exist an element $b \in Q$ such that $f(b) =a$. substituting $a$ with $f(b_0)$ in the definition you get that $f(b_0) \in Q$ then $b_0 \in Q$ taking the contrapositive of this statement you get if $b_0 \notin Q$ then $f(b_0) \notin Q$. ($a \implies b$ if and only if $\neg b \implies \neg a$).
In general for any function $f:A\to B$ you do not have this property a very trivial example is a function $f:\{0,1\}\to \{0\}$ We have that $f(\{0\})-f(\{1\})=\emptyset$ but $f(\{0\}-\{1\})=f(\{0\})=\{0\}$