Explaination of a justification: additive functors preserve limits












3












$begingroup$



Lemma: For any collection ${ M_i}_{iin I}$ of $R$-modules, and $R$-module $N$, there is a natural isomorphism
$${rm Hom}_R(oplus_i M_i, N)cong prod_i {rm Hom}_R(M_i,N).$$
Proof: Additive functors preserve limits.




[Ref: this link, page 2.]



Q. The lemma also follows from the definition of direct sum of modules. However, the purely categorical justification given in above proof is not clear to me. Can one explain in detail the above proof?





I just started study of homological algebra, so my vocabulary of this sunbect is not so deep.










share|cite|improve this question











$endgroup$












  • $begingroup$
    "Additive functors preserve limits" is completely false; the author probably meant "products" rather "limits".
    $endgroup$
    – Arnaud D.
    Apr 3 at 5:21










  • $begingroup$
    I added link; I do not know what author wants to say in one line and if it is correct. (Also, I am very beginner in this subject also. Really, I do not know technicalities.)
    $endgroup$
    – Beginner
    Apr 3 at 5:32






  • 1




    $begingroup$
    @ArnaudD.: Additive functors don't preserve (possibly infinite) products either, though!
    $endgroup$
    – Eric Wofsey
    Apr 3 at 6:41










  • $begingroup$
    @EricWofsey Indeed! Derek's answer makes me wonder if perhaps the author meant "representable functors"...
    $endgroup$
    – Arnaud D.
    Apr 3 at 6:43
















3












$begingroup$



Lemma: For any collection ${ M_i}_{iin I}$ of $R$-modules, and $R$-module $N$, there is a natural isomorphism
$${rm Hom}_R(oplus_i M_i, N)cong prod_i {rm Hom}_R(M_i,N).$$
Proof: Additive functors preserve limits.




[Ref: this link, page 2.]



Q. The lemma also follows from the definition of direct sum of modules. However, the purely categorical justification given in above proof is not clear to me. Can one explain in detail the above proof?





I just started study of homological algebra, so my vocabulary of this sunbect is not so deep.










share|cite|improve this question











$endgroup$












  • $begingroup$
    "Additive functors preserve limits" is completely false; the author probably meant "products" rather "limits".
    $endgroup$
    – Arnaud D.
    Apr 3 at 5:21










  • $begingroup$
    I added link; I do not know what author wants to say in one line and if it is correct. (Also, I am very beginner in this subject also. Really, I do not know technicalities.)
    $endgroup$
    – Beginner
    Apr 3 at 5:32






  • 1




    $begingroup$
    @ArnaudD.: Additive functors don't preserve (possibly infinite) products either, though!
    $endgroup$
    – Eric Wofsey
    Apr 3 at 6:41










  • $begingroup$
    @EricWofsey Indeed! Derek's answer makes me wonder if perhaps the author meant "representable functors"...
    $endgroup$
    – Arnaud D.
    Apr 3 at 6:43














3












3








3





$begingroup$



Lemma: For any collection ${ M_i}_{iin I}$ of $R$-modules, and $R$-module $N$, there is a natural isomorphism
$${rm Hom}_R(oplus_i M_i, N)cong prod_i {rm Hom}_R(M_i,N).$$
Proof: Additive functors preserve limits.




[Ref: this link, page 2.]



Q. The lemma also follows from the definition of direct sum of modules. However, the purely categorical justification given in above proof is not clear to me. Can one explain in detail the above proof?





I just started study of homological algebra, so my vocabulary of this sunbect is not so deep.










share|cite|improve this question











$endgroup$





Lemma: For any collection ${ M_i}_{iin I}$ of $R$-modules, and $R$-module $N$, there is a natural isomorphism
$${rm Hom}_R(oplus_i M_i, N)cong prod_i {rm Hom}_R(M_i,N).$$
Proof: Additive functors preserve limits.




[Ref: this link, page 2.]



Q. The lemma also follows from the definition of direct sum of modules. However, the purely categorical justification given in above proof is not clear to me. Can one explain in detail the above proof?





I just started study of homological algebra, so my vocabulary of this sunbect is not so deep.







abstract-algebra modules homological-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 3 at 5:30







Beginner

















asked Apr 3 at 4:59









BeginnerBeginner

4,10811226




4,10811226












  • $begingroup$
    "Additive functors preserve limits" is completely false; the author probably meant "products" rather "limits".
    $endgroup$
    – Arnaud D.
    Apr 3 at 5:21










  • $begingroup$
    I added link; I do not know what author wants to say in one line and if it is correct. (Also, I am very beginner in this subject also. Really, I do not know technicalities.)
    $endgroup$
    – Beginner
    Apr 3 at 5:32






  • 1




    $begingroup$
    @ArnaudD.: Additive functors don't preserve (possibly infinite) products either, though!
    $endgroup$
    – Eric Wofsey
    Apr 3 at 6:41










  • $begingroup$
    @EricWofsey Indeed! Derek's answer makes me wonder if perhaps the author meant "representable functors"...
    $endgroup$
    – Arnaud D.
    Apr 3 at 6:43


















  • $begingroup$
    "Additive functors preserve limits" is completely false; the author probably meant "products" rather "limits".
    $endgroup$
    – Arnaud D.
    Apr 3 at 5:21










  • $begingroup$
    I added link; I do not know what author wants to say in one line and if it is correct. (Also, I am very beginner in this subject also. Really, I do not know technicalities.)
    $endgroup$
    – Beginner
    Apr 3 at 5:32






  • 1




    $begingroup$
    @ArnaudD.: Additive functors don't preserve (possibly infinite) products either, though!
    $endgroup$
    – Eric Wofsey
    Apr 3 at 6:41










  • $begingroup$
    @EricWofsey Indeed! Derek's answer makes me wonder if perhaps the author meant "representable functors"...
    $endgroup$
    – Arnaud D.
    Apr 3 at 6:43
















$begingroup$
"Additive functors preserve limits" is completely false; the author probably meant "products" rather "limits".
$endgroup$
– Arnaud D.
Apr 3 at 5:21




$begingroup$
"Additive functors preserve limits" is completely false; the author probably meant "products" rather "limits".
$endgroup$
– Arnaud D.
Apr 3 at 5:21












$begingroup$
I added link; I do not know what author wants to say in one line and if it is correct. (Also, I am very beginner in this subject also. Really, I do not know technicalities.)
$endgroup$
– Beginner
Apr 3 at 5:32




$begingroup$
I added link; I do not know what author wants to say in one line and if it is correct. (Also, I am very beginner in this subject also. Really, I do not know technicalities.)
$endgroup$
– Beginner
Apr 3 at 5:32




1




1




$begingroup$
@ArnaudD.: Additive functors don't preserve (possibly infinite) products either, though!
$endgroup$
– Eric Wofsey
Apr 3 at 6:41




$begingroup$
@ArnaudD.: Additive functors don't preserve (possibly infinite) products either, though!
$endgroup$
– Eric Wofsey
Apr 3 at 6:41












$begingroup$
@EricWofsey Indeed! Derek's answer makes me wonder if perhaps the author meant "representable functors"...
$endgroup$
– Arnaud D.
Apr 3 at 6:43




$begingroup$
@EricWofsey Indeed! Derek's answer makes me wonder if perhaps the author meant "representable functors"...
$endgroup$
– Arnaud D.
Apr 3 at 6:43










2 Answers
2






active

oldest

votes


















5












$begingroup$

If you know that the functor $T=operatorname{Hom}_R(-,M):Rmathtt{Mod}^{op}to mathtt{Ab}$ preserves limits, then the result follows. Indeed, the direct sum $bigoplus M_i$ is just the coproduct of the $M_i$ in $Rmathtt{Mod}$ and thus the product of the $M_i$ in $Rmathtt{Mod}^{op}$, and so since $T$ preserves limits the natural map $T(bigoplus M_i)to prod T(M_i)$ is an isomorphism, and this exactly gives the statement of the Lemma.



However, the justification given in the proof of the Lemma is totally wrong. Additive functors do not always preserve limits, and so you cannot use additivity of $T$ to deduce that it preserves limits. It turns out that $T$ does preserve limits, but you must prove this by other means (and given that the Lemma is just a special case of $T$ preserving limits, this doesn't actually make proving the Lemma any easier).






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    Here's a purely categorical approach when $R$ is commutative. (The proof below actually works for any symmetric monoidally closed category which $Rtext{Mod}$ is in the commutative $R$ case.)



    The internal (or external) Hom is continuous in both arguments. Since it's contravariant in the first argument, this looks like turning colimits to limits. For the external Hom, this is basically just the perspective of limits and colimits by representability, so it could even be true by definition. For the internal Hom, we need to be working in a symmetric monoidally closed category (with the monoidal closure as the Hom). The symmetry is important here. Symmetry gives us the adjunction $$mathsf{Hom}^{op}(mathsf{Hom}_R(N,P),M)=mathsf{Hom}(M,mathsf{Hom}_R(N,P))congmathsf{Hom}(N,mathsf{Hom}_R(M,P))$$ (natural in $M$ and $N$ [and $P$]) meaning $mathsf{Hom}_R(-,P)$ is a right adjoint and thus preserves limits like all right adjoints.






    share|cite|improve this answer











    $endgroup$









    • 2




      $begingroup$
      $R$ seems to not be assumed to be commutative though, so this is not actually an internal Hom.
      $endgroup$
      – Eric Wofsey
      Apr 3 at 6:42










    • $begingroup$
      @EricWofsey You're right. I've added commutativity of $R$ as an assumption for now.
      $endgroup$
      – Derek Elkins
      Apr 3 at 6:55












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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    If you know that the functor $T=operatorname{Hom}_R(-,M):Rmathtt{Mod}^{op}to mathtt{Ab}$ preserves limits, then the result follows. Indeed, the direct sum $bigoplus M_i$ is just the coproduct of the $M_i$ in $Rmathtt{Mod}$ and thus the product of the $M_i$ in $Rmathtt{Mod}^{op}$, and so since $T$ preserves limits the natural map $T(bigoplus M_i)to prod T(M_i)$ is an isomorphism, and this exactly gives the statement of the Lemma.



    However, the justification given in the proof of the Lemma is totally wrong. Additive functors do not always preserve limits, and so you cannot use additivity of $T$ to deduce that it preserves limits. It turns out that $T$ does preserve limits, but you must prove this by other means (and given that the Lemma is just a special case of $T$ preserving limits, this doesn't actually make proving the Lemma any easier).






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      If you know that the functor $T=operatorname{Hom}_R(-,M):Rmathtt{Mod}^{op}to mathtt{Ab}$ preserves limits, then the result follows. Indeed, the direct sum $bigoplus M_i$ is just the coproduct of the $M_i$ in $Rmathtt{Mod}$ and thus the product of the $M_i$ in $Rmathtt{Mod}^{op}$, and so since $T$ preserves limits the natural map $T(bigoplus M_i)to prod T(M_i)$ is an isomorphism, and this exactly gives the statement of the Lemma.



      However, the justification given in the proof of the Lemma is totally wrong. Additive functors do not always preserve limits, and so you cannot use additivity of $T$ to deduce that it preserves limits. It turns out that $T$ does preserve limits, but you must prove this by other means (and given that the Lemma is just a special case of $T$ preserving limits, this doesn't actually make proving the Lemma any easier).






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        If you know that the functor $T=operatorname{Hom}_R(-,M):Rmathtt{Mod}^{op}to mathtt{Ab}$ preserves limits, then the result follows. Indeed, the direct sum $bigoplus M_i$ is just the coproduct of the $M_i$ in $Rmathtt{Mod}$ and thus the product of the $M_i$ in $Rmathtt{Mod}^{op}$, and so since $T$ preserves limits the natural map $T(bigoplus M_i)to prod T(M_i)$ is an isomorphism, and this exactly gives the statement of the Lemma.



        However, the justification given in the proof of the Lemma is totally wrong. Additive functors do not always preserve limits, and so you cannot use additivity of $T$ to deduce that it preserves limits. It turns out that $T$ does preserve limits, but you must prove this by other means (and given that the Lemma is just a special case of $T$ preserving limits, this doesn't actually make proving the Lemma any easier).






        share|cite|improve this answer









        $endgroup$



        If you know that the functor $T=operatorname{Hom}_R(-,M):Rmathtt{Mod}^{op}to mathtt{Ab}$ preserves limits, then the result follows. Indeed, the direct sum $bigoplus M_i$ is just the coproduct of the $M_i$ in $Rmathtt{Mod}$ and thus the product of the $M_i$ in $Rmathtt{Mod}^{op}$, and so since $T$ preserves limits the natural map $T(bigoplus M_i)to prod T(M_i)$ is an isomorphism, and this exactly gives the statement of the Lemma.



        However, the justification given in the proof of the Lemma is totally wrong. Additive functors do not always preserve limits, and so you cannot use additivity of $T$ to deduce that it preserves limits. It turns out that $T$ does preserve limits, but you must prove this by other means (and given that the Lemma is just a special case of $T$ preserving limits, this doesn't actually make proving the Lemma any easier).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 3 at 6:40









        Eric WofseyEric Wofsey

        193k14221352




        193k14221352























            3












            $begingroup$

            Here's a purely categorical approach when $R$ is commutative. (The proof below actually works for any symmetric monoidally closed category which $Rtext{Mod}$ is in the commutative $R$ case.)



            The internal (or external) Hom is continuous in both arguments. Since it's contravariant in the first argument, this looks like turning colimits to limits. For the external Hom, this is basically just the perspective of limits and colimits by representability, so it could even be true by definition. For the internal Hom, we need to be working in a symmetric monoidally closed category (with the monoidal closure as the Hom). The symmetry is important here. Symmetry gives us the adjunction $$mathsf{Hom}^{op}(mathsf{Hom}_R(N,P),M)=mathsf{Hom}(M,mathsf{Hom}_R(N,P))congmathsf{Hom}(N,mathsf{Hom}_R(M,P))$$ (natural in $M$ and $N$ [and $P$]) meaning $mathsf{Hom}_R(-,P)$ is a right adjoint and thus preserves limits like all right adjoints.






            share|cite|improve this answer











            $endgroup$









            • 2




              $begingroup$
              $R$ seems to not be assumed to be commutative though, so this is not actually an internal Hom.
              $endgroup$
              – Eric Wofsey
              Apr 3 at 6:42










            • $begingroup$
              @EricWofsey You're right. I've added commutativity of $R$ as an assumption for now.
              $endgroup$
              – Derek Elkins
              Apr 3 at 6:55
















            3












            $begingroup$

            Here's a purely categorical approach when $R$ is commutative. (The proof below actually works for any symmetric monoidally closed category which $Rtext{Mod}$ is in the commutative $R$ case.)



            The internal (or external) Hom is continuous in both arguments. Since it's contravariant in the first argument, this looks like turning colimits to limits. For the external Hom, this is basically just the perspective of limits and colimits by representability, so it could even be true by definition. For the internal Hom, we need to be working in a symmetric monoidally closed category (with the monoidal closure as the Hom). The symmetry is important here. Symmetry gives us the adjunction $$mathsf{Hom}^{op}(mathsf{Hom}_R(N,P),M)=mathsf{Hom}(M,mathsf{Hom}_R(N,P))congmathsf{Hom}(N,mathsf{Hom}_R(M,P))$$ (natural in $M$ and $N$ [and $P$]) meaning $mathsf{Hom}_R(-,P)$ is a right adjoint and thus preserves limits like all right adjoints.






            share|cite|improve this answer











            $endgroup$









            • 2




              $begingroup$
              $R$ seems to not be assumed to be commutative though, so this is not actually an internal Hom.
              $endgroup$
              – Eric Wofsey
              Apr 3 at 6:42










            • $begingroup$
              @EricWofsey You're right. I've added commutativity of $R$ as an assumption for now.
              $endgroup$
              – Derek Elkins
              Apr 3 at 6:55














            3












            3








            3





            $begingroup$

            Here's a purely categorical approach when $R$ is commutative. (The proof below actually works for any symmetric monoidally closed category which $Rtext{Mod}$ is in the commutative $R$ case.)



            The internal (or external) Hom is continuous in both arguments. Since it's contravariant in the first argument, this looks like turning colimits to limits. For the external Hom, this is basically just the perspective of limits and colimits by representability, so it could even be true by definition. For the internal Hom, we need to be working in a symmetric monoidally closed category (with the monoidal closure as the Hom). The symmetry is important here. Symmetry gives us the adjunction $$mathsf{Hom}^{op}(mathsf{Hom}_R(N,P),M)=mathsf{Hom}(M,mathsf{Hom}_R(N,P))congmathsf{Hom}(N,mathsf{Hom}_R(M,P))$$ (natural in $M$ and $N$ [and $P$]) meaning $mathsf{Hom}_R(-,P)$ is a right adjoint and thus preserves limits like all right adjoints.






            share|cite|improve this answer











            $endgroup$



            Here's a purely categorical approach when $R$ is commutative. (The proof below actually works for any symmetric monoidally closed category which $Rtext{Mod}$ is in the commutative $R$ case.)



            The internal (or external) Hom is continuous in both arguments. Since it's contravariant in the first argument, this looks like turning colimits to limits. For the external Hom, this is basically just the perspective of limits and colimits by representability, so it could even be true by definition. For the internal Hom, we need to be working in a symmetric monoidally closed category (with the monoidal closure as the Hom). The symmetry is important here. Symmetry gives us the adjunction $$mathsf{Hom}^{op}(mathsf{Hom}_R(N,P),M)=mathsf{Hom}(M,mathsf{Hom}_R(N,P))congmathsf{Hom}(N,mathsf{Hom}_R(M,P))$$ (natural in $M$ and $N$ [and $P$]) meaning $mathsf{Hom}_R(-,P)$ is a right adjoint and thus preserves limits like all right adjoints.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Apr 3 at 6:53

























            answered Apr 3 at 6:29









            Derek ElkinsDerek Elkins

            17.7k11437




            17.7k11437








            • 2




              $begingroup$
              $R$ seems to not be assumed to be commutative though, so this is not actually an internal Hom.
              $endgroup$
              – Eric Wofsey
              Apr 3 at 6:42










            • $begingroup$
              @EricWofsey You're right. I've added commutativity of $R$ as an assumption for now.
              $endgroup$
              – Derek Elkins
              Apr 3 at 6:55














            • 2




              $begingroup$
              $R$ seems to not be assumed to be commutative though, so this is not actually an internal Hom.
              $endgroup$
              – Eric Wofsey
              Apr 3 at 6:42










            • $begingroup$
              @EricWofsey You're right. I've added commutativity of $R$ as an assumption for now.
              $endgroup$
              – Derek Elkins
              Apr 3 at 6:55








            2




            2




            $begingroup$
            $R$ seems to not be assumed to be commutative though, so this is not actually an internal Hom.
            $endgroup$
            – Eric Wofsey
            Apr 3 at 6:42




            $begingroup$
            $R$ seems to not be assumed to be commutative though, so this is not actually an internal Hom.
            $endgroup$
            – Eric Wofsey
            Apr 3 at 6:42












            $begingroup$
            @EricWofsey You're right. I've added commutativity of $R$ as an assumption for now.
            $endgroup$
            – Derek Elkins
            Apr 3 at 6:55




            $begingroup$
            @EricWofsey You're right. I've added commutativity of $R$ as an assumption for now.
            $endgroup$
            – Derek Elkins
            Apr 3 at 6:55


















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