Why is the composition of a surjective and injective function neither surjective nor injective?












0














I am currently preparing for an exam coming up and I was therefore looking in earlier exam-sets, and found a question with a solution I just cannot make sense in my head.




Let $A, B, C$ be three arbitrary sets, $fcolon Bto C$ be a surjective function and $gcolon A to B$ be an injective function. Then the composition $fcirc g$ is?




  1. Injective

  2. Surjective

  3. Bijectie

  4. Neither injective nor surjective




The answer-sheet suggests that the composition of these two functions should neither be injective nor surjective. Whenever I try drawing functions by example however, it seems easy to make a composition that is both injective and surjective.



Thanks.










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  • 4




    The problem is poorly worded. The correct answer is "we can't make any conclusions".
    – MathematicsStudent1122
    2 days ago
















0














I am currently preparing for an exam coming up and I was therefore looking in earlier exam-sets, and found a question with a solution I just cannot make sense in my head.




Let $A, B, C$ be three arbitrary sets, $fcolon Bto C$ be a surjective function and $gcolon A to B$ be an injective function. Then the composition $fcirc g$ is?




  1. Injective

  2. Surjective

  3. Bijectie

  4. Neither injective nor surjective




The answer-sheet suggests that the composition of these two functions should neither be injective nor surjective. Whenever I try drawing functions by example however, it seems easy to make a composition that is both injective and surjective.



Thanks.










share|cite|improve this question









New contributor




Carrazo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 4




    The problem is poorly worded. The correct answer is "we can't make any conclusions".
    – MathematicsStudent1122
    2 days ago














0












0








0







I am currently preparing for an exam coming up and I was therefore looking in earlier exam-sets, and found a question with a solution I just cannot make sense in my head.




Let $A, B, C$ be three arbitrary sets, $fcolon Bto C$ be a surjective function and $gcolon A to B$ be an injective function. Then the composition $fcirc g$ is?




  1. Injective

  2. Surjective

  3. Bijectie

  4. Neither injective nor surjective




The answer-sheet suggests that the composition of these two functions should neither be injective nor surjective. Whenever I try drawing functions by example however, it seems easy to make a composition that is both injective and surjective.



Thanks.










share|cite|improve this question









New contributor




Carrazo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I am currently preparing for an exam coming up and I was therefore looking in earlier exam-sets, and found a question with a solution I just cannot make sense in my head.




Let $A, B, C$ be three arbitrary sets, $fcolon Bto C$ be a surjective function and $gcolon A to B$ be an injective function. Then the composition $fcirc g$ is?




  1. Injective

  2. Surjective

  3. Bijectie

  4. Neither injective nor surjective




The answer-sheet suggests that the composition of these two functions should neither be injective nor surjective. Whenever I try drawing functions by example however, it seems easy to make a composition that is both injective and surjective.



Thanks.







functions elementary-set-theory function-and-relation-composition






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Carrazo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question









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edited 2 days ago









Asaf Karagila

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asked 2 days ago









Carrazo

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Carrazo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.








  • 4




    The problem is poorly worded. The correct answer is "we can't make any conclusions".
    – MathematicsStudent1122
    2 days ago














  • 4




    The problem is poorly worded. The correct answer is "we can't make any conclusions".
    – MathematicsStudent1122
    2 days ago








4




4




The problem is poorly worded. The correct answer is "we can't make any conclusions".
– MathematicsStudent1122
2 days ago




The problem is poorly worded. The correct answer is "we can't make any conclusions".
– MathematicsStudent1122
2 days ago










3 Answers
3






active

oldest

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2














Of course you can construct examples where the composition is injective, or where it is surjective. But the point is that you can also construct examples where neither of these properties is fulfilled. The question has to be interpreted in the way "Can one conclude from $f$ being surjective and $g$ being injective that the composition is necessarily surjective, etc.?'' And this is not the case.






share|cite|improve this answer

















  • 1




    Thank you for the answer, but would you then agree, following that logic, that we cannot conclude either that the composition of the two functions necessarily is not injective nor surjective?
    – Carrazo
    2 days ago






  • 1




    @Carrazo Yes, I agree. The problem is wrongly formulated.
    – Gerhard S.
    2 days ago



















1














The answer to the question, as it is written, should probably be "There is not enough information to conclude anything about the injectivity or surjectivity of the composition." However, as this is clearly not the intended answer (as it is not an option in your list of multiple choices), we have to get into the heads of the question writers and try to figure out what they meant to ask. My reading of the question is that it would better be phrased as




Let $A$, $B$, $C$ be three arbitrary sets, $f : B to C$ be a surjective function and $g : A to B$ be an injective function. Which of the following statements is true:





  1. $gcirc f$ is always injective.


  2. $gcirc f$ is always surjective.


  3. $gcirc f$ is always bijective.




In this phrasing, the correct answer is that all three statements are false. Again, there isn't enough information to conclude anything about the surjectiveity or injectivity of the composition. To see this, we need only construct two counterexamples:




  • Let $A = {ast}$ be a singleton set, and let $B = C = { a, b }$ be a set with two elements. Define $g : A to B$ by $g(ast) = a$, and take $f : B to C$ to be the identity map. Then the composition $fcirc g$ is injective (any map from a singleton set to any other set must be injective), but is not surjective ($b$ is not the image of any point in $A$ under the composition). Therefore the composition needn't be surjective (which also means that it needn't be bijective, as every bijective map is necessarily surjective).


  • Let $A = B = {a, b}$ be a set with two elements, and $C = {ast}$ be a singleton set. Let $f : Ato B$ be the identity map, and let $g : Bto C$ be the only possible map, i.e. the function defined by $g(x) = ast$ for all $xin B$. Then the composition $fcirc g$ is surjective, but not injective. Therefore the composition needn't be injective.



Note that we can construct examples of sets $A$, $B$, and $C$, and maps $f : Ato B$ and $g : B to C$ such that the composition $fcirc g$ is injective (but not surjective), surjective (but not injective), or bijective. However, simply knowing that $f$ is injective and that $g$ is surjective is not enough to tell us anything about the properties of the composition.






share|cite|improve this answer





























    0














    You can make the composition both injective and surjective (the easiest example is to take $f$ and $g$ to both be the identity map on $A = B = C$). The point is that you can't conclude, from the data provided, that it's either. The easy way to see this is to have one of $f$ and $g$ be the identity, and the other either surjective but not injective or injective but not surjective: then the composition is just the same as the non-identity one, which doesn't have the relevant property.






    share|cite|improve this answer





















    • I can understand that we cannot necessarily conclude from what is provided that it is either injective or surjective. But how can we conclude that it in fact is neither? Excuse me if I'm just completely misunderstanding the question.
      – Carrazo
      2 days ago










    • You can't, see my example. Indeed, even if $f$ is not injective and $g$ is not surjective, you still can't conclude that: take $V$ a vector space and $U$ a proper subspace, take $f$ to be the projection from $V$ to $U$, and take $g$ to be the inclusion from $U$ to $V$. Then $f$ is surjective but not injective, $g$ is injective but not surjective, and $fcirc g$ is the identity on $U$, so is bijective.
      – user3482749
      2 days ago











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    3 Answers
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    3 Answers
    3






    active

    oldest

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    active

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    active

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    2














    Of course you can construct examples where the composition is injective, or where it is surjective. But the point is that you can also construct examples where neither of these properties is fulfilled. The question has to be interpreted in the way "Can one conclude from $f$ being surjective and $g$ being injective that the composition is necessarily surjective, etc.?'' And this is not the case.






    share|cite|improve this answer

















    • 1




      Thank you for the answer, but would you then agree, following that logic, that we cannot conclude either that the composition of the two functions necessarily is not injective nor surjective?
      – Carrazo
      2 days ago






    • 1




      @Carrazo Yes, I agree. The problem is wrongly formulated.
      – Gerhard S.
      2 days ago
















    2














    Of course you can construct examples where the composition is injective, or where it is surjective. But the point is that you can also construct examples where neither of these properties is fulfilled. The question has to be interpreted in the way "Can one conclude from $f$ being surjective and $g$ being injective that the composition is necessarily surjective, etc.?'' And this is not the case.






    share|cite|improve this answer

















    • 1




      Thank you for the answer, but would you then agree, following that logic, that we cannot conclude either that the composition of the two functions necessarily is not injective nor surjective?
      – Carrazo
      2 days ago






    • 1




      @Carrazo Yes, I agree. The problem is wrongly formulated.
      – Gerhard S.
      2 days ago














    2












    2








    2






    Of course you can construct examples where the composition is injective, or where it is surjective. But the point is that you can also construct examples where neither of these properties is fulfilled. The question has to be interpreted in the way "Can one conclude from $f$ being surjective and $g$ being injective that the composition is necessarily surjective, etc.?'' And this is not the case.






    share|cite|improve this answer












    Of course you can construct examples where the composition is injective, or where it is surjective. But the point is that you can also construct examples where neither of these properties is fulfilled. The question has to be interpreted in the way "Can one conclude from $f$ being surjective and $g$ being injective that the composition is necessarily surjective, etc.?'' And this is not the case.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 days ago









    Gerhard S.

    99529




    99529








    • 1




      Thank you for the answer, but would you then agree, following that logic, that we cannot conclude either that the composition of the two functions necessarily is not injective nor surjective?
      – Carrazo
      2 days ago






    • 1




      @Carrazo Yes, I agree. The problem is wrongly formulated.
      – Gerhard S.
      2 days ago














    • 1




      Thank you for the answer, but would you then agree, following that logic, that we cannot conclude either that the composition of the two functions necessarily is not injective nor surjective?
      – Carrazo
      2 days ago






    • 1




      @Carrazo Yes, I agree. The problem is wrongly formulated.
      – Gerhard S.
      2 days ago








    1




    1




    Thank you for the answer, but would you then agree, following that logic, that we cannot conclude either that the composition of the two functions necessarily is not injective nor surjective?
    – Carrazo
    2 days ago




    Thank you for the answer, but would you then agree, following that logic, that we cannot conclude either that the composition of the two functions necessarily is not injective nor surjective?
    – Carrazo
    2 days ago




    1




    1




    @Carrazo Yes, I agree. The problem is wrongly formulated.
    – Gerhard S.
    2 days ago




    @Carrazo Yes, I agree. The problem is wrongly formulated.
    – Gerhard S.
    2 days ago











    1














    The answer to the question, as it is written, should probably be "There is not enough information to conclude anything about the injectivity or surjectivity of the composition." However, as this is clearly not the intended answer (as it is not an option in your list of multiple choices), we have to get into the heads of the question writers and try to figure out what they meant to ask. My reading of the question is that it would better be phrased as




    Let $A$, $B$, $C$ be three arbitrary sets, $f : B to C$ be a surjective function and $g : A to B$ be an injective function. Which of the following statements is true:





    1. $gcirc f$ is always injective.


    2. $gcirc f$ is always surjective.


    3. $gcirc f$ is always bijective.




    In this phrasing, the correct answer is that all three statements are false. Again, there isn't enough information to conclude anything about the surjectiveity or injectivity of the composition. To see this, we need only construct two counterexamples:




    • Let $A = {ast}$ be a singleton set, and let $B = C = { a, b }$ be a set with two elements. Define $g : A to B$ by $g(ast) = a$, and take $f : B to C$ to be the identity map. Then the composition $fcirc g$ is injective (any map from a singleton set to any other set must be injective), but is not surjective ($b$ is not the image of any point in $A$ under the composition). Therefore the composition needn't be surjective (which also means that it needn't be bijective, as every bijective map is necessarily surjective).


    • Let $A = B = {a, b}$ be a set with two elements, and $C = {ast}$ be a singleton set. Let $f : Ato B$ be the identity map, and let $g : Bto C$ be the only possible map, i.e. the function defined by $g(x) = ast$ for all $xin B$. Then the composition $fcirc g$ is surjective, but not injective. Therefore the composition needn't be injective.



    Note that we can construct examples of sets $A$, $B$, and $C$, and maps $f : Ato B$ and $g : B to C$ such that the composition $fcirc g$ is injective (but not surjective), surjective (but not injective), or bijective. However, simply knowing that $f$ is injective and that $g$ is surjective is not enough to tell us anything about the properties of the composition.






    share|cite|improve this answer


























      1














      The answer to the question, as it is written, should probably be "There is not enough information to conclude anything about the injectivity or surjectivity of the composition." However, as this is clearly not the intended answer (as it is not an option in your list of multiple choices), we have to get into the heads of the question writers and try to figure out what they meant to ask. My reading of the question is that it would better be phrased as




      Let $A$, $B$, $C$ be three arbitrary sets, $f : B to C$ be a surjective function and $g : A to B$ be an injective function. Which of the following statements is true:





      1. $gcirc f$ is always injective.


      2. $gcirc f$ is always surjective.


      3. $gcirc f$ is always bijective.




      In this phrasing, the correct answer is that all three statements are false. Again, there isn't enough information to conclude anything about the surjectiveity or injectivity of the composition. To see this, we need only construct two counterexamples:




      • Let $A = {ast}$ be a singleton set, and let $B = C = { a, b }$ be a set with two elements. Define $g : A to B$ by $g(ast) = a$, and take $f : B to C$ to be the identity map. Then the composition $fcirc g$ is injective (any map from a singleton set to any other set must be injective), but is not surjective ($b$ is not the image of any point in $A$ under the composition). Therefore the composition needn't be surjective (which also means that it needn't be bijective, as every bijective map is necessarily surjective).


      • Let $A = B = {a, b}$ be a set with two elements, and $C = {ast}$ be a singleton set. Let $f : Ato B$ be the identity map, and let $g : Bto C$ be the only possible map, i.e. the function defined by $g(x) = ast$ for all $xin B$. Then the composition $fcirc g$ is surjective, but not injective. Therefore the composition needn't be injective.



      Note that we can construct examples of sets $A$, $B$, and $C$, and maps $f : Ato B$ and $g : B to C$ such that the composition $fcirc g$ is injective (but not surjective), surjective (but not injective), or bijective. However, simply knowing that $f$ is injective and that $g$ is surjective is not enough to tell us anything about the properties of the composition.






      share|cite|improve this answer
























        1












        1








        1






        The answer to the question, as it is written, should probably be "There is not enough information to conclude anything about the injectivity or surjectivity of the composition." However, as this is clearly not the intended answer (as it is not an option in your list of multiple choices), we have to get into the heads of the question writers and try to figure out what they meant to ask. My reading of the question is that it would better be phrased as




        Let $A$, $B$, $C$ be three arbitrary sets, $f : B to C$ be a surjective function and $g : A to B$ be an injective function. Which of the following statements is true:





        1. $gcirc f$ is always injective.


        2. $gcirc f$ is always surjective.


        3. $gcirc f$ is always bijective.




        In this phrasing, the correct answer is that all three statements are false. Again, there isn't enough information to conclude anything about the surjectiveity or injectivity of the composition. To see this, we need only construct two counterexamples:




        • Let $A = {ast}$ be a singleton set, and let $B = C = { a, b }$ be a set with two elements. Define $g : A to B$ by $g(ast) = a$, and take $f : B to C$ to be the identity map. Then the composition $fcirc g$ is injective (any map from a singleton set to any other set must be injective), but is not surjective ($b$ is not the image of any point in $A$ under the composition). Therefore the composition needn't be surjective (which also means that it needn't be bijective, as every bijective map is necessarily surjective).


        • Let $A = B = {a, b}$ be a set with two elements, and $C = {ast}$ be a singleton set. Let $f : Ato B$ be the identity map, and let $g : Bto C$ be the only possible map, i.e. the function defined by $g(x) = ast$ for all $xin B$. Then the composition $fcirc g$ is surjective, but not injective. Therefore the composition needn't be injective.



        Note that we can construct examples of sets $A$, $B$, and $C$, and maps $f : Ato B$ and $g : B to C$ such that the composition $fcirc g$ is injective (but not surjective), surjective (but not injective), or bijective. However, simply knowing that $f$ is injective and that $g$ is surjective is not enough to tell us anything about the properties of the composition.






        share|cite|improve this answer












        The answer to the question, as it is written, should probably be "There is not enough information to conclude anything about the injectivity or surjectivity of the composition." However, as this is clearly not the intended answer (as it is not an option in your list of multiple choices), we have to get into the heads of the question writers and try to figure out what they meant to ask. My reading of the question is that it would better be phrased as




        Let $A$, $B$, $C$ be three arbitrary sets, $f : B to C$ be a surjective function and $g : A to B$ be an injective function. Which of the following statements is true:





        1. $gcirc f$ is always injective.


        2. $gcirc f$ is always surjective.


        3. $gcirc f$ is always bijective.




        In this phrasing, the correct answer is that all three statements are false. Again, there isn't enough information to conclude anything about the surjectiveity or injectivity of the composition. To see this, we need only construct two counterexamples:




        • Let $A = {ast}$ be a singleton set, and let $B = C = { a, b }$ be a set with two elements. Define $g : A to B$ by $g(ast) = a$, and take $f : B to C$ to be the identity map. Then the composition $fcirc g$ is injective (any map from a singleton set to any other set must be injective), but is not surjective ($b$ is not the image of any point in $A$ under the composition). Therefore the composition needn't be surjective (which also means that it needn't be bijective, as every bijective map is necessarily surjective).


        • Let $A = B = {a, b}$ be a set with two elements, and $C = {ast}$ be a singleton set. Let $f : Ato B$ be the identity map, and let $g : Bto C$ be the only possible map, i.e. the function defined by $g(x) = ast$ for all $xin B$. Then the composition $fcirc g$ is surjective, but not injective. Therefore the composition needn't be injective.



        Note that we can construct examples of sets $A$, $B$, and $C$, and maps $f : Ato B$ and $g : B to C$ such that the composition $fcirc g$ is injective (but not surjective), surjective (but not injective), or bijective. However, simply knowing that $f$ is injective and that $g$ is surjective is not enough to tell us anything about the properties of the composition.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Xander Henderson

        14.1k103554




        14.1k103554























            0














            You can make the composition both injective and surjective (the easiest example is to take $f$ and $g$ to both be the identity map on $A = B = C$). The point is that you can't conclude, from the data provided, that it's either. The easy way to see this is to have one of $f$ and $g$ be the identity, and the other either surjective but not injective or injective but not surjective: then the composition is just the same as the non-identity one, which doesn't have the relevant property.






            share|cite|improve this answer





















            • I can understand that we cannot necessarily conclude from what is provided that it is either injective or surjective. But how can we conclude that it in fact is neither? Excuse me if I'm just completely misunderstanding the question.
              – Carrazo
              2 days ago










            • You can't, see my example. Indeed, even if $f$ is not injective and $g$ is not surjective, you still can't conclude that: take $V$ a vector space and $U$ a proper subspace, take $f$ to be the projection from $V$ to $U$, and take $g$ to be the inclusion from $U$ to $V$. Then $f$ is surjective but not injective, $g$ is injective but not surjective, and $fcirc g$ is the identity on $U$, so is bijective.
              – user3482749
              2 days ago
















            0














            You can make the composition both injective and surjective (the easiest example is to take $f$ and $g$ to both be the identity map on $A = B = C$). The point is that you can't conclude, from the data provided, that it's either. The easy way to see this is to have one of $f$ and $g$ be the identity, and the other either surjective but not injective or injective but not surjective: then the composition is just the same as the non-identity one, which doesn't have the relevant property.






            share|cite|improve this answer





















            • I can understand that we cannot necessarily conclude from what is provided that it is either injective or surjective. But how can we conclude that it in fact is neither? Excuse me if I'm just completely misunderstanding the question.
              – Carrazo
              2 days ago










            • You can't, see my example. Indeed, even if $f$ is not injective and $g$ is not surjective, you still can't conclude that: take $V$ a vector space and $U$ a proper subspace, take $f$ to be the projection from $V$ to $U$, and take $g$ to be the inclusion from $U$ to $V$. Then $f$ is surjective but not injective, $g$ is injective but not surjective, and $fcirc g$ is the identity on $U$, so is bijective.
              – user3482749
              2 days ago














            0












            0








            0






            You can make the composition both injective and surjective (the easiest example is to take $f$ and $g$ to both be the identity map on $A = B = C$). The point is that you can't conclude, from the data provided, that it's either. The easy way to see this is to have one of $f$ and $g$ be the identity, and the other either surjective but not injective or injective but not surjective: then the composition is just the same as the non-identity one, which doesn't have the relevant property.






            share|cite|improve this answer












            You can make the composition both injective and surjective (the easiest example is to take $f$ and $g$ to both be the identity map on $A = B = C$). The point is that you can't conclude, from the data provided, that it's either. The easy way to see this is to have one of $f$ and $g$ be the identity, and the other either surjective but not injective or injective but not surjective: then the composition is just the same as the non-identity one, which doesn't have the relevant property.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 days ago









            user3482749

            2,753414




            2,753414












            • I can understand that we cannot necessarily conclude from what is provided that it is either injective or surjective. But how can we conclude that it in fact is neither? Excuse me if I'm just completely misunderstanding the question.
              – Carrazo
              2 days ago










            • You can't, see my example. Indeed, even if $f$ is not injective and $g$ is not surjective, you still can't conclude that: take $V$ a vector space and $U$ a proper subspace, take $f$ to be the projection from $V$ to $U$, and take $g$ to be the inclusion from $U$ to $V$. Then $f$ is surjective but not injective, $g$ is injective but not surjective, and $fcirc g$ is the identity on $U$, so is bijective.
              – user3482749
              2 days ago


















            • I can understand that we cannot necessarily conclude from what is provided that it is either injective or surjective. But how can we conclude that it in fact is neither? Excuse me if I'm just completely misunderstanding the question.
              – Carrazo
              2 days ago










            • You can't, see my example. Indeed, even if $f$ is not injective and $g$ is not surjective, you still can't conclude that: take $V$ a vector space and $U$ a proper subspace, take $f$ to be the projection from $V$ to $U$, and take $g$ to be the inclusion from $U$ to $V$. Then $f$ is surjective but not injective, $g$ is injective but not surjective, and $fcirc g$ is the identity on $U$, so is bijective.
              – user3482749
              2 days ago
















            I can understand that we cannot necessarily conclude from what is provided that it is either injective or surjective. But how can we conclude that it in fact is neither? Excuse me if I'm just completely misunderstanding the question.
            – Carrazo
            2 days ago




            I can understand that we cannot necessarily conclude from what is provided that it is either injective or surjective. But how can we conclude that it in fact is neither? Excuse me if I'm just completely misunderstanding the question.
            – Carrazo
            2 days ago












            You can't, see my example. Indeed, even if $f$ is not injective and $g$ is not surjective, you still can't conclude that: take $V$ a vector space and $U$ a proper subspace, take $f$ to be the projection from $V$ to $U$, and take $g$ to be the inclusion from $U$ to $V$. Then $f$ is surjective but not injective, $g$ is injective but not surjective, and $fcirc g$ is the identity on $U$, so is bijective.
            – user3482749
            2 days ago




            You can't, see my example. Indeed, even if $f$ is not injective and $g$ is not surjective, you still can't conclude that: take $V$ a vector space and $U$ a proper subspace, take $f$ to be the projection from $V$ to $U$, and take $g$ to be the inclusion from $U$ to $V$. Then $f$ is surjective but not injective, $g$ is injective but not surjective, and $fcirc g$ is the identity on $U$, so is bijective.
            – user3482749
            2 days ago










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