I’m having trouble proving the following theorem. I’ll show my proof so far. I would really appreciate if you could help me, so I can improve my proof and correct some mistakes.
Theorem: Let $f:A \rightarrow B$ be a map. Think of this map as inducing the map $f_{*}:\mathcal{P}(A) \rightarrow \mathcal{P}(B)$, and the map $f^{*}:\mathcal{P}(B) \rightarrow \mathcal{P}(A)$. Then, $f_{*}$ is surjective if and only if $f$ is surjective.
Proof: $\implies.$ Suppose that $f_{*}$ is surjective. Then, for all $Y \subseteq B$, there exists some $X \subseteq A$, such that $f_{*}(X)=Y$. In particular, for $Y=B$, there is some set $X \subseteq A$, such that $f_{*}(X)=B$. Let $X_0$ be that set. Since $X_{0} \subseteq A$, we know that $f_{*}(X_{0}) \subseteq f_{*}(A)$. Observe that $f_{*}(A) \subseteq B$. Given that $f_{*}(X_{0}) = B$, then $f_{*}(X_{0}) \subseteq B$. From $f_{*}(X_{0}) \subseteq f_{*}(A)$ we conclude that $B \subseteq f_{*}(A)$. So we have that $f_{*}(A)=B$, therefore $f$ is surjective.
$\Longleftarrow.$ Suppose that $f$ is surjective. From a former theorem, I know that $F = f_{*}(f^{*}(F))$ for all $F \subseteq B$. Let $S \subseteq B$ be any set. Observe that $S \in \mathcal{P}(B)$. We define the set $R \subseteq A$ as $R=f^{*}(S)$. Note that $R \in \mathcal{P}(A)$. So $f_{*}(R) = f_{*}(f^{*}(S)) = S$. Therefore $f_{*}$ is surjective. $\square$
Is this proof right? I think this could be better and I want to improve it! Any help is welcome. Thank you in advance!
Best Answer
The "$\implies$" part is very well written, but you introduced $X_0$ and kept writing it as $X$
For the "$\Longleftarrow$" part, $f_*(f^*(R))$ is not defined, since $R\in \mathcal P(A)$, and nothing implies that $R\in \mathcal P(B)$ or $f_{*}(f^{*}(R)) = f_{*}(f^{*}(S))$ . However, correcting this detail and writing $f_*(R)=f_*(f^*(S))=S$ will do.
It seems that the only mistakes are in the notation, and besides that the proof is ok.