I wish to prove the existence of an injective function $g:B\to A$ given a surjective function $f:A\to B$. This sounds simple enough, however I'm having trouble writing a formal proof for it.
Thanks in advance.
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3Actually this result is a direct consequence of the axiom of choice (I suspect that it is equivalent). Do you know it? Do you know how to apply it? – Crostul Aug 29 '15 at 13:03
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@OscarZegarra I think your statement is exactly the formulation of axiom of choice, so there is nothing to 'prove' here. – user2097 Aug 29 '15 at 13:09
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1It is equivalent to AC, – DanielWainfleet Aug 29 '15 at 13:12
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See also: There exists an injection from $X$ to $Y$ if and only if there exists a surjection from $Y$ to $X$. – Martin Sleziak Feb 05 '22 at 10:14
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For each $b \in B$ let $F(b)$ be the subset of $A$ such that $f(a) =b$ for all $a \in F(b)$.
Then since $f$ is surjective each $F(b)$ is non empty.
Axiom of choice says I can form a set $F'$ by picking one element from each $F(b)$.
Now define $g: B \to A$ by $g(b) = a$ where $f(a) = b$ with $a \in F'$
Tom Collinge
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for $b\in{B}$ since $f$ issurjective $f^{-1}(b)$ isnot empty. using the axiom of choice $g(b)\in{f^{-1}(b)}$.
R.N
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The Axiom of Choice (AC): If $ F$ is a set of non-empty sets then there is a function $g$ with $dom(g)=F$ such that $ g(x) \in f$ for each $x \in F$ . Such a function $g$ is called a choice function for $ F$. So given a surjection $f:A \to B$, the axioms of Separation and Comprehension give us $ F$ , the set of fibers of $f$, that is $ F=\{f^{-1} \{ b \} : b \in B \}$ where $f^{-1} \{b\} \}= \{ a \in A : f(a)=b \}$. So apply AC...... I suspect that the converse also holds: That your Q implies AC.
DanielWainfleet
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Maybe it's clearer to say that if $F=\bigcup_{\alpha \in \Lambda } F_{\alpha }$ then there is a function $g:\Lambda \to F$ such that $g(\alpha )\in F_{\alpha }$ – Matematleta Aug 29 '15 at 14:40