Suppose it weren't necessary that f is injective. Then there would exist a non-injective function f such that for any two subsets A,B⊆X we get f(A∩B)=f(A)∩f(B).
Let a,b∈X
f({a})∩f({b})={f(a)=f(b)}=f({a}∩{b})=f(∅)=∅
Contradiction.
Suppose it weren't necessary that f is surjective.
Now assume a∈f−1({a}). Then by definition of inverse image of set, f(a)∈{a} meaning a=f(a). Contradiction.
''injective + surjective = 1-1''
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