Proving |A|<=Aleph Null with Function f:A->B and |B|<=Aleph Null

[MathematicalPhysicist]

i have a function f:A->B, I am also given that |B|<=null aleph, and for every b in B, |f^-1({b})|<=null aleph, i need to prove that |A|<=null aleph.
basically i think that A equals the union of f^-1({b}) for every b in B, and by another theorem i can consequently assert that |A|<=null aleph.

but is this correct?

 

[AKG]

Yes...

 

[MathematicalPhysicist]

AKG, thanks.