With the axiom of choice this is trivial, but is there any way to construct this injection explicitly in the ZF system?
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Don't you mean an injection $B\to A$? If so then it is equivalent with AC. – drhab Apr 28 '20 at 10:18
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I don't. As you say, that's equivalent to the axiom of choice. This is provable without the axiom of choice. – Banbadle Apr 28 '20 at 10:49
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Okay. My distrust was triggered by the fact that your question has a quite "natural" solution. – drhab Apr 28 '20 at 10:51
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With the axiom of choice, then it's not ZF, but ZFC. – Asaf Karagila Apr 28 '20 at 13:08
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I know, I was specifying without the axiom of choice. The question you've marked this as a duplicate of is not the same. – Banbadle Apr 28 '20 at 20:40
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No need to even mention Choice here, you can just use $f$ describe the surjection explicitly (and don't call that "$f$" too!) – BrianO May 03 '20 at 18:02
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What about $g\colon B\rightarrow\mathcal{P}(A),\,b\mapsto f^{-1}(\{b\})$. Suppose that there are $b,b^{\prime}\in B$ with $g(b)=g(b^{\prime})$. This means $f^{-1}(\{b\})=f^{-1}(\{b^{\prime}\})$. The surjectivity of $f$ implies that neither of these sets is empty, hence there is an $a\in f^{-1}(\{b\})=f^{-1}(\{b^{\prime}\})$, meaning $f(a)=b$ and $f(a)=b^{\prime}$, i.e. $b=b^{\prime}$. Therefore, $g$ is an injection.
Thorgott
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That'll do it, thanks. I'm annoyed, I wrote this down and decided it wasn't necessarily injective. – Banbadle Apr 28 '20 at 10:23
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Function $g:B\to\mathcal P(A)$ prescribed by: $$b\mapsto f^{-1}(\{b\})=\{a\in A\mid f(a)=b\}$$ is injective if function $f:A\to B$ is surjective.
(and also if $B-f(A)$ is a singleton, so surjectivity of $f$ cannot even be classified as a necessary condition).
drhab
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