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Statement: For every surjective function $f:A \to B$ there exists set $C\subseteq A$ such that function $f:C \to B$ is bijection.

As I see it, this is obviously true for finite sets, in way that for every multiple occurrence of some element in $B$, it is possible to just eliminate all elements in $A$ whose image is that specific element in $B$ but one. However I am not sure about infinite sets, as I can't quite come up with counterexample, if there is one.

Thanks for help.

N. F. Taussig
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Dovla
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    There's nothing in your argument that depends on anything being finite (with the possible exception of dependience on the Axiom of Choice, which you need pay no attention to at this stage of your studies). – Ethan Bolker Nov 19 '18 at 12:57

1 Answers1

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If $f:A\rightarrow B$ is a surjective mapping, then the set of preimages $\{f^{-1}(b)\mid b\in B\}$ with $f^{-1}(b) = \{a\in A\mid f(a)=b\}$ forms a partition of $A$. Now choose for each set $f^{-1}(b)$ one representative $a_b\in A$ with $f(a_b)=b$ and you are done. See Bolker's comment.

Wuestenfux
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    +1 Correct and well said, but I suspect the OP hasn't learned about partitions (and equivalence relations) yet. – Ethan Bolker Nov 19 '18 at 13:14
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    Do we have that the restricted funcion is bijective because the domain and the range has the same number of elements? @Wuestenfux – Mary Star Nov 19 '20 at 21:08