Proving f is onto given if $g circ f = h circ f$, then g = h












2












$begingroup$


I realize that there has already been an answer to this problem. But I want to know if my proof was correct. Thank you for your time.



Problem



Suppose A,B, and C are sets and $f: A rightarrow B$



Suppose that C has at least two elements, and for all functions $g$ and $h$ from B to C, if $g circ h = h circ f$ then $g = h$. Prove that $f$ is onto.



Proof



Suppose f is not onto. Then there is $b_1$ such that for all $a in A$, $f(a) neq b_1$. Suppose $(b_1, c_1) in g$ and $(b_1, c_2) in h$. For the assumption that $g circ h = h circ f$ then $g = h$ to be true, $(b_1, c_1) = (b_1, c_2)$ whenever $g circ f = h circ f $.



Assume $g circ f = h circ f$. Since $b_1 notin Ran(f)$, $(a,c_1) notin g circ f$ and $(a,c_2) notin h circ f$. This means that as long as there are at least two elements in C, it is possible that $(b_1, c_2) neq (b_1, c_2)$ while $g circ f = h circ f$. (Even if $c_1 neq c_2$, it can still be true that $g circ f = h circ f$ since $c_1$ and $c_2$ are not in the range of $g circ f$ and $h circ f$.)



This is a contradiction, hence $f$ is onto.






functions proof-verification elementary-set-theory proof-writing function-and-relation-composition







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I realize that there has already been an answer to this problem. But I want to know if my proof was correct. Thank you for your time.



    Problem



    Suppose A,B, and C are sets and $f: A rightarrow B$



    Suppose that C has at least two elements, and for all functions $g$ and $h$ from B to C, if $g circ h = h circ f$ then $g = h$. Prove that $f$ is onto.



    Proof



    Suppose f is not onto. Then there is $b_1$ such that for all $a in A$, $f(a) neq b_1$. Suppose $(b_1, c_1) in g$ and $(b_1, c_2) in h$. For the assumption that $g circ h = h circ f$ then $g = h$ to be true, $(b_1, c_1) = (b_1, c_2)$ whenever $g circ f = h circ f $.



    Assume $g circ f = h circ f$. Since $b_1 notin Ran(f)$, $(a,c_1) notin g circ f$ and $(a,c_2) notin h circ f$. This means that as long as there are at least two elements in C, it is possible that $(b_1, c_2) neq (b_1, c_2)$ while $g circ f = h circ f$. (Even if $c_1 neq c_2$, it can still be true that $g circ f = h circ f$ since $c_1$ and $c_2$ are not in the range of $g circ f$ and $h circ f$.)



    This is a contradiction, hence $f$ is onto.






    functions proof-verification elementary-set-theory proof-writing function-and-relation-composition







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      I realize that there has already been an answer to this problem. But I want to know if my proof was correct. Thank you for your time.



      Problem



      Suppose A,B, and C are sets and $f: A rightarrow B$



      Suppose that C has at least two elements, and for all functions $g$ and $h$ from B to C, if $g circ h = h circ f$ then $g = h$. Prove that $f$ is onto.



      Proof



      Suppose f is not onto. Then there is $b_1$ such that for all $a in A$, $f(a) neq b_1$. Suppose $(b_1, c_1) in g$ and $(b_1, c_2) in h$. For the assumption that $g circ h = h circ f$ then $g = h$ to be true, $(b_1, c_1) = (b_1, c_2)$ whenever $g circ f = h circ f $.



      Assume $g circ f = h circ f$. Since $b_1 notin Ran(f)$, $(a,c_1) notin g circ f$ and $(a,c_2) notin h circ f$. This means that as long as there are at least two elements in C, it is possible that $(b_1, c_2) neq (b_1, c_2)$ while $g circ f = h circ f$. (Even if $c_1 neq c_2$, it can still be true that $g circ f = h circ f$ since $c_1$ and $c_2$ are not in the range of $g circ f$ and $h circ f$.)



      This is a contradiction, hence $f$ is onto.






      functions proof-verification elementary-set-theory proof-writing function-and-relation-composition







      share|cite|improve this question











      $endgroup$




      I realize that there has already been an answer to this problem. But I want to know if my proof was correct. Thank you for your time.



      Problem



      Suppose A,B, and C are sets and $f: A rightarrow B$



      Suppose that C has at least two elements, and for all functions $g$ and $h$ from B to C, if $g circ h = h circ f$ then $g = h$. Prove that $f$ is onto.



      Proof



      Suppose f is not onto. Then there is $b_1$ such that for all $a in A$, $f(a) neq b_1$. Suppose $(b_1, c_1) in g$ and $(b_1, c_2) in h$. For the assumption that $g circ h = h circ f$ then $g = h$ to be true, $(b_1, c_1) = (b_1, c_2)$ whenever $g circ f = h circ f $.



      Assume $g circ f = h circ f$. Since $b_1 notin Ran(f)$, $(a,c_1) notin g circ f$ and $(a,c_2) notin h circ f$. This means that as long as there are at least two elements in C, it is possible that $(b_1, c_2) neq (b_1, c_2)$ while $g circ f = h circ f$. (Even if $c_1 neq c_2$, it can still be true that $g circ f = h circ f$ since $c_1$ and $c_2$ are not in the range of $g circ f$ and $h circ f$.)



      This is a contradiction, hence $f$ is onto.






      functions proof-verification elementary-set-theory proof-writing function-and-relation-composition




      functions proof-verification elementary-set-theory proof-writing function-and-relation-composition






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Dec 12 '18 at 1:39









      Martin Sleziak

      44.9k10122277




      44.9k10122277










      asked Sep 22 '18 at 18:57









      PhilPhil

      4716




      4716






















          1 Answer
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          $begingroup$

          After the end of the second sentence, your first paragraph is a mess. There is no $g$ or $h$ defined or assumed yet, so you can’t talk about them, and certainly not talk about pairs in $g$ or in $h$. The final sentence “For the assumption...” is also incorrect as written and does not make much sense. That whole thing needs to be removed.



          Second paragraph also starts badly. Who are $c_1$ and $c_2$? Who are $g$ and $h$? They are not given, so you can’t take them as given. Finally, any ‘proof’ that ends with “It’s possible that” or “it can still be true” is not really a proof. You aren’t proving something, you are just suggesting that it might be the case that something happens. That’s not a proof. In a proof, you are supposed to show that something does happen, unequivocally. Until you actually exhibit it happening, you are not proving anything.



          So I would say that this attempt is not correct.



          What you want to do after your first sentence is to actually construct two functions $g,hcolon Bto C$ with the properties that $gcirc f = hcirc f$, and $gneq h$. That will give a you proof by contrapositive, rather than one by contradiction.






          share|cite|improve this answer









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            0












            $begingroup$

            After the end of the second sentence, your first paragraph is a mess. There is no $g$ or $h$ defined or assumed yet, so you can’t talk about them, and certainly not talk about pairs in $g$ or in $h$. The final sentence “For the assumption...” is also incorrect as written and does not make much sense. That whole thing needs to be removed.



            Second paragraph also starts badly. Who are $c_1$ and $c_2$? Who are $g$ and $h$? They are not given, so you can’t take them as given. Finally, any ‘proof’ that ends with “It’s possible that” or “it can still be true” is not really a proof. You aren’t proving something, you are just suggesting that it might be the case that something happens. That’s not a proof. In a proof, you are supposed to show that something does happen, unequivocally. Until you actually exhibit it happening, you are not proving anything.



            So I would say that this attempt is not correct.



            What you want to do after your first sentence is to actually construct two functions $g,hcolon Bto C$ with the properties that $gcirc f = hcirc f$, and $gneq h$. That will give a you proof by contrapositive, rather than one by contradiction.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              After the end of the second sentence, your first paragraph is a mess. There is no $g$ or $h$ defined or assumed yet, so you can’t talk about them, and certainly not talk about pairs in $g$ or in $h$. The final sentence “For the assumption...” is also incorrect as written and does not make much sense. That whole thing needs to be removed.



              Second paragraph also starts badly. Who are $c_1$ and $c_2$? Who are $g$ and $h$? They are not given, so you can’t take them as given. Finally, any ‘proof’ that ends with “It’s possible that” or “it can still be true” is not really a proof. You aren’t proving something, you are just suggesting that it might be the case that something happens. That’s not a proof. In a proof, you are supposed to show that something does happen, unequivocally. Until you actually exhibit it happening, you are not proving anything.



              So I would say that this attempt is not correct.



              What you want to do after your first sentence is to actually construct two functions $g,hcolon Bto C$ with the properties that $gcirc f = hcirc f$, and $gneq h$. That will give a you proof by contrapositive, rather than one by contradiction.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                After the end of the second sentence, your first paragraph is a mess. There is no $g$ or $h$ defined or assumed yet, so you can’t talk about them, and certainly not talk about pairs in $g$ or in $h$. The final sentence “For the assumption...” is also incorrect as written and does not make much sense. That whole thing needs to be removed.



                Second paragraph also starts badly. Who are $c_1$ and $c_2$? Who are $g$ and $h$? They are not given, so you can’t take them as given. Finally, any ‘proof’ that ends with “It’s possible that” or “it can still be true” is not really a proof. You aren’t proving something, you are just suggesting that it might be the case that something happens. That’s not a proof. In a proof, you are supposed to show that something does happen, unequivocally. Until you actually exhibit it happening, you are not proving anything.



                So I would say that this attempt is not correct.



                What you want to do after your first sentence is to actually construct two functions $g,hcolon Bto C$ with the properties that $gcirc f = hcirc f$, and $gneq h$. That will give a you proof by contrapositive, rather than one by contradiction.






                share|cite|improve this answer









                $endgroup$



                After the end of the second sentence, your first paragraph is a mess. There is no $g$ or $h$ defined or assumed yet, so you can’t talk about them, and certainly not talk about pairs in $g$ or in $h$. The final sentence “For the assumption...” is also incorrect as written and does not make much sense. That whole thing needs to be removed.



                Second paragraph also starts badly. Who are $c_1$ and $c_2$? Who are $g$ and $h$? They are not given, so you can’t take them as given. Finally, any ‘proof’ that ends with “It’s possible that” or “it can still be true” is not really a proof. You aren’t proving something, you are just suggesting that it might be the case that something happens. That’s not a proof. In a proof, you are supposed to show that something does happen, unequivocally. Until you actually exhibit it happening, you are not proving anything.



                So I would say that this attempt is not correct.



                What you want to do after your first sentence is to actually construct two functions $g,hcolon Bto C$ with the properties that $gcirc f = hcirc f$, and $gneq h$. That will give a you proof by contrapositive, rather than one by contradiction.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 22 '18 at 20:01









                Arturo MagidinArturo Magidin

                266k34590920




                266k34590920






























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