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I'm struggling on how to prove this statement:

Let $X,Y$ two sets and $f : X \to Y$ and $g: P(X)\to P(Y)$ two functions s.t. $g(A) = f(A)$. Prove that $f$ is bijective $\iff$ $g$ is bijective.

So what I did so far is proving that if one is injective, the another one is as well. I did this:

$f$ is injective $\iff f(A)=f(B) \implies A=B\implies g(A) = f(A) = f(B)=g(B)\implies g$ is injective because $g(A) = f(A)$.

Now what is left to show is that if one is surjective, the other one is. But I don't know how to approach there.

lambdar
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    Did you realize that $g(A)$ and $f(A)$ have different meanings? $g(A)$ means the value of the function $g$ evaluated at $A$, while $f(A)$ means the set ${f(a): a \in A}$. – WhatsUp Oct 30 '21 at 22:55
  • @WhatsUp So I need to take an element from $A$ and see how it goes with $f$? – lambdar Oct 30 '21 at 23:43

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