Restricting domain of surjective function to make bijection











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Statement:
For every surjective function $f:A to B$ there exists set $Csubseteq A$ 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.






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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
    yesterday

















up vote
2
down vote

favorite












Statement:
For every surjective function $f:A to B$ there exists set $Csubseteq A$ 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.






functions







share|cite|improve this question




















  • 2




    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
    yesterday















up vote
2
down vote

favorite









up vote
2
down vote

favorite











Statement:
For every surjective function $f:A to B$ there exists set $Csubseteq A$ 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.






functions







share|cite|improve this question















Statement:
For every surjective function $f:A to B$ there exists set $Csubseteq A$ 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.






functions




functions






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edited yesterday









N. F. Taussig

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asked yesterday









Dovla

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  • 2




    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
    yesterday
















  • 2




    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
    yesterday










2




2




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
yesterday






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
yesterday












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






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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
    yesterday











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






share|cite|improve this answer

















  • 1




    +1 Correct and well said, but I suspect the OP hasn't learned about partitions (and equivalence relations) yet.
    – Ethan Bolker
    yesterday















up vote
2
down vote













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






share|cite|improve this answer

















  • 1




    +1 Correct and well said, but I suspect the OP hasn't learned about partitions (and equivalence relations) yet.
    – Ethan Bolker
    yesterday













up vote
2
down vote










up vote
2
down vote









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






share|cite|improve this answer












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







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share|cite|improve this answer



share|cite|improve this answer










answered yesterday









Wuestenfux

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2,3441410








  • 1




    +1 Correct and well said, but I suspect the OP hasn't learned about partitions (and equivalence relations) yet.
    – Ethan Bolker
    yesterday














  • 1




    +1 Correct and well said, but I suspect the OP hasn't learned about partitions (and equivalence relations) yet.
    – Ethan Bolker
    yesterday








1




1




+1 Correct and well said, but I suspect the OP hasn't learned about partitions (and equivalence relations) yet.
– Ethan Bolker
yesterday




+1 Correct and well said, but I suspect the OP hasn't learned about partitions (and equivalence relations) yet.
– Ethan Bolker
yesterday


















 

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