Why is f being an injection equivalent to this ?

[fatineouahbi]

1.Prove f(A⋂B) ⊂ f(A) ⋂ f(B)
2.Prove f(A) ⋂ f(B) ⊂ f(A⋂B) ⟺ f is an injection

I've solved the first question , as for the second I started with f(A) ⋂ f(B) ⊂ f(A⋂B) ⇒ f is an injection this way :
Let's suppose f(a) = f(b) = p
If we consider A = {a} and B = {b} then f(A) = f(B) = p
then f(A) ⋂ f(B) = p
then f(A⋂B) = p (because f(A) ⋂ f(B) ⊂ f(A⋂B) from the supposition and f(A⋂B) ⊂ f(A) ⋂ f(B) from the first question)
then A⋂B ≠ ∅
then a=b
then f is an injection .

But I don't know how to solve "f is an injection ⇒ f(A) ⋂ f(B) ⊂ f(A⋂B)"

 

[Ackbach]

$(\implies)$ Assume $f(A)\cap f(B) \subset f(A\cap B)$ for all sets $A$ and $B$. Suppose $A=\{x\}$ and $B=\{y\}$ are singleton sets. Further assume that $f(x)=f(y)$. Then $f(A)=f(B)$, so $f(A)\cap f(B)$ is nonempty. By assumption, $f(A\cap B)$ is nonempty, since $f(A)\cap f(B)\subset f(A\cap B)$. This implies $A\cap B$ is nonempty. Since $A$ and $B$ are singletons, this implies $A=B$, or $\{x\}=\{y\}$, or $x=y$. Hence, $f$ is an injection.

$(\impliedby)$ This is a "what do you know" kind of proof. Can you write it out?