Proving: $f$ is injective $\Leftrightarrow f(X \cap Y) = f(X) \cap f(Y)$ [duplicate]

Let $A$ and $B$ be sets, where $f : A \rightarrow B$ is a function. Show that the following properties are valid equivalent*:

1. $f$ is injective.
2. For all $X, Y \subset A$ is valid: $f(X \cap Y)=f(X)\cap f(Y)$
3. For all $X \subset Y \subset A$ is valid: $f(X \setminus Y)=f(X) \setminus f(Y)$.

I do know what injective is, but I thought number (2.) and (3.) were valid for any kind of function. Just to see if I understood this right:

$f(X\cap Y)$ means, first, make the intersection from $X$ and $Y$ and then map it to $B$ via $f$.

$f(X)\cap f(Y)$ actually means, map all $f(X)$ and $f(Y)$ and intersect both.

Aren't those both properties valid for all functions? I can't think of a counter example. Thanks in advance guys!

*edited by SN.

No they are not true for any function:

Take a function $f:\{0,1\}\to\{0,1\}$ such that $f(0)=1$ and $f(1)=1$. Then $f[\{0\}\cap\{1\}]=f[\varnothing]=\varnothing$ but $f[\{0\}]\cap f[\{1\}]=\{1\}\cap\{1\}=\{1\}$. This function provides a counterexample for the second case as well: $f[\{0,1\}\setminus\{0\}]=\{1\}$, while $f[\{0,1\}]\setminus f[\{0\}]=\{1\}\setminus\{1\}=\varnothing$.

Note that for any function $f$ it is true that $f[X\cap Y]\subseteq f[X]\cap f[Y]$ and it is also true that $f[X]\setminus f[Y]\subseteq f[X\setminus Y]$.

As for the equivalence, a function $f$ is called injective exactly when $x\neq y$ implies $f(x)\neq f(y)$ (or equivalently $f(x)=f(y)$ implies $x=y$). They are sometimes called $1-1$:

$1\Rightarrow 2$. Let $f:A\to B$ be injective. We just need to show that $f[X]\cap f[Y]\subseteq f[X\cap Y]$. Let $x\in f[X]\cap f[Y]$. Then there is some $a\in X$ and some $b\in Y$ such that $f(a)=f(b)=x$. By the definition of injective functions $a=b$ thus $a\in X\cap Y$ or $x\in f[X\cap Y]$.

$2\Rightarrow 1$. Now let $f[X]\cap f[Y]=f[X\cap Y]$. Let $f(a)=f(b)$. We have that $f[\{a\}\cap\{b\}]=f[\{a\}]\cap f[\{b\}]$. Thus $f[\{a\}\cap\{b\}]$ is not empty (since it is equal to $f[\{a\}]\cap f[\{b\}]$ which is not). Therefore $\{a\}\cap\{b\}$ is not empty, which means that $a=b$.

$1\Rightarrow 3$. Let $f:A\to B$ be injective. We just need to show that $f[X\setminus Y]\subseteq f[X]\setminus f[Y]$. Let $x\in f[X\setminus Y]$. Of course $x\in f[X]$. We have that there is some $a\in X\setminus Y$ such that $f(a)=x$. For every $b\in Y$ we have that $a\neq b$ thus $f(a)\neq f(b)$. Thus $x\notin f[Y]$ and thus $x\in f[X]\setminus f[Y]$.

$3\Rightarrow 1$. Conversely assume that $f[X\setminus Y]= f[X]\setminus f[Y]$. Let $f(a)=f(b)$. Then $f[\{a,b\}\setminus\{b\}]=f[\{a,b\}]\setminus f[\{b\}]$. The second set is empty, thus $f[\{a,b\}\setminus\{b\}]$ is empty. Then $\{a,b\}\setminus\{b\}$ is empty, which means $a=b$.