Is $f$ a one-to-one function on $S$?

[Dustinsfl]

Let $f:S\to T$ be a function.
Prove that the following statements are equivalent.a
$f$ is one-to-one on $S$.

b
$f(A\cap B) = f(A)\cap f(B)$ for all subsets $A,B$ of $S$.

c
$f^{-1}[f(A)] = A$ for every subset $A$ of $S$.

d
For all disjoint subsets $A$ and $B$ of $S$, the images $f(A)$ and $f(B)$ are disjoint.

Having a tough time with this one.

 

[Plato2]

Let $f:S\to T$ be a function.
Prove that the following statements are equivalent.
a
$f$ is one-to-one on $S$.

b
$f(A\cap B) = f(A)\cap f(B)$ for all subsets $A,B$ of $S$.

For any function $f$ we have $f(A\cap B)\subseteq f(A)\cap f(B)$.
So suppose that $f$ is one-to-one and $t\in f(A)\cap f(B)$.
$\left( {\exists {a_t} \in A} \right)\left[ {f({a_t}) = t} \right]~\&~\left( {\exists {b_t} \in B} \right)\left[ {f({b_t}) = t} \right]$.
Use one-to-one to prove $t\in f(A\cap B)$.

Now start with $f(A\cap B) = f(A)\cap f(B)$ and continue.

 

[Dustinsfl]

For any function $f$ we have $f(A\cap B)\subseteq f(A)\cap f(B)$.
So suppose that $f$ is one-to-one and $t\in f(A)\cap f(B)$.
$\left( {\exists {a_t} \in A} \right)\left[ {f({a_t}) = t} \right]~\&~\left( {\exists {b_t} \in B} \right)\left[ {f({b_t}) = t} \right]$.
Use one-to-one to prove $t\in f(A\cap B)$.

Now start with $f(A\cap B) = f(A)\cap f(B)$ and continue.

How can $f(A\cap B) = f(A)\cap f(B)$ be used to show c is true? I don't get it.

 

[Plato2]

How can $f(A\cap B) = f(A)\cap f(B)$ be used to show c is true? I don't get it.

I would show that $a) \Leftrightarrow c)$.

 

[Evgeny.Makarov]

Note that $f^{-1}(f(A))\supseteq A$, so one only has to show $f^{-1}(f(A))\subseteq A$. Suppose $f^{-1}(f(A))=A\cup B$ where $A\cap B=\emptyset$. Using (b) show that $f(B)=\emptyset$ and therefore $B=\emptyset$.