Question #117328
Prove the following result by contradiction:
Let f : X TO Y be a mapping. Suppose f (A INTERSECTION B) = f (A) INTERSECTION f (B) for all subsets
A, B PROPER SET X , f (PHI) = PHI. Then f is a 1-1 mapping.
1
Expert's answer
2020-05-21T14:47:00-0400

 Suppose it weren't necessary that ff is injective. Then there would exist a non-injective function ff such that for any two subsets A,BXA,B⊆X we get f(AB)=f(A)f(B)f(A∩B)=f(A)∩f(B).

Let a,bXa,b\in X

f({a})f({b})={f(a)=f(b)}f({a}{b})=f()=f(\{a\})\cap f(\{b\})=\{f(a)=f(b)\}\ne f(\{a\}\cap\{b\})\\ =f(\emptyset)=\emptyset

Contradiction.


Suppose it weren't necessary that ff is surjective.

Now assume a∉f1({a})a \not\in f^{−1}(\{a\}). Then by definition of inverse image of set, f(a)∉{a}f(a) \not\in \{a\} meaning af(a)a\ne f(a). Contradiction.


''injective + surjective = 1-1''


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