$f^{*}$ is surjective if and only if $f$ is injective

I’m having a hard time understanding this proof and I hope someone could help me.

Theorem: Let $f: A rightarrow B$ a map. Think of this map as inducing the map $f^{*}: mathcal{P}(B) rightarrow mathcal{P}(A)$. Then, $f^{*}$ is surjective if and only if $f$ is injective.

The $Longleftarrow$ part I already prove it:

Proof: $Longleftarrow.$ Suppose $f$ is injective. Hence, we know that $E = f^{*}(f_{*}(E))$ for all subsets $E subseteq A$. Let $S$ be a subset of $A$. Then $S in mathcal{P}(A)$. We define the set $X_0$ as $X_0 = f_{*}(S)$. Observe that $X_0 in mathcal{P}(B)$. Hence $f^{*}(X_0) = f^{*}(f_{*}(S)) = S$. Therefore $f^{*}$ is surjective. $square$

For the $implies$ part, I don’t know what to do.

My attempts:

  1. I tried to prove the contrapositive, so if $f$ is not injective, then $f^{*}$ is not surjective. Suppose that $f$ is not injective. Then there exists some $a,b in A$ such that $a neq b$ and $f(a) = f(b)$. But I don’t know what to do next.

  2. I tried a direct proof: Suppose that $f^{*}$ is surjective. Hence for all $X in mathcal{P}(A)$, there exists some $Y in mathcal{P}(B)$, such that $f^{*}(Y)=X$. Since $A subseteq A$, we have that $A in mathcal{P}(A)$. So there exists some $Y_0 in mathcal{P}(B)$ such that $f^{*}(Y_0) = A$. But, again, I don’t know where to go next.

Can someone help me? Thank you in advance!

Mathematics Asked by Air Mike on November 14, 2021

2 Answers

2 Answers

Following on from your attempt 1. You need to show that ${a}$ is not in the image of $f^*$. Specifically, show that if $ain f^*(X)$ then ${a}neq f^*(X)$.

Answered by tkf on November 14, 2021

Suppose $x,y$ are two distinct elements of $A$ such that $f(x)=f(y)$. Since $f^*$ is surjective there is a subset $Esubseteq B$ such that ${x}=f^*(E)$. Now, we have $xin{x}=f^*(E)$ and hence $f(x)in E$. Thus:

$f(y)=f(x)in E$

But this means $yin f^*(E)={x}$, a contradiction.

Answered by Mark on November 14, 2021

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