# $f^{*}$ 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

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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