Explaination of a justification: additive functors preserve limitsAdditive functors preserve split exact sequencesshowing exact functors preserve exact sequences (abelian categories, additive functors, and kernels)Morita contexts without tearsQuick Question on a Proof of Artin-Wedderburn TheoremAdditive exact functors preserve homology of modulesThe direct limit of morphisms and the direct limit of tensor product functors$Pcong P^ast$ iff $P$ is a f.g projective module?Modules, Direct Sums, ConfusionAdditive functors preserve direct summandsAdditive Functors preserve Null Sequences

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Explaination of a justification: additive functors preserve limits


Additive functors preserve split exact sequencesshowing exact functors preserve exact sequences (abelian categories, additive functors, and kernels)Morita contexts without tearsQuick Question on a Proof of Artin-Wedderburn TheoremAdditive exact functors preserve homology of modulesThe direct limit of morphisms and the direct limit of tensor product functors$Pcong P^ast$ iff $P$ is a f.g projective module?Modules, Direct Sums, ConfusionAdditive functors preserve direct summandsAdditive Functors preserve Null Sequences













2












$begingroup$



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.
    yesterday










  • $begingroup$
    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
    yesterday






  • 1




    $begingroup$
    @ArnaudD.: Additive functors don't preserve (possibly infinite) products either, though!
    $endgroup$
    – Eric Wofsey
    23 hours ago










  • $begingroup$
    @EricWofsey Indeed! Derek's answer makes me wonder if perhaps the author meant "representable functors"...
    $endgroup$
    – Arnaud D.
    23 hours ago















2












$begingroup$



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.
    yesterday










  • $begingroup$
    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
    yesterday






  • 1




    $begingroup$
    @ArnaudD.: Additive functors don't preserve (possibly infinite) products either, though!
    $endgroup$
    – Eric Wofsey
    23 hours ago










  • $begingroup$
    @EricWofsey Indeed! Derek's answer makes me wonder if perhaps the author meant "representable functors"...
    $endgroup$
    – Arnaud D.
    23 hours ago













2












2








2





$begingroup$



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 yesterday







Beginner

















asked yesterday









BeginnerBeginner

4,08511226




4,08511226











  • $begingroup$
    "Additive functors preserve limits" is completely false; the author probably meant "products" rather "limits".
    $endgroup$
    – Arnaud D.
    yesterday










  • $begingroup$
    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
    yesterday






  • 1




    $begingroup$
    @ArnaudD.: Additive functors don't preserve (possibly infinite) products either, though!
    $endgroup$
    – Eric Wofsey
    23 hours ago










  • $begingroup$
    @EricWofsey Indeed! Derek's answer makes me wonder if perhaps the author meant "representable functors"...
    $endgroup$
    – Arnaud D.
    23 hours ago
















  • $begingroup$
    "Additive functors preserve limits" is completely false; the author probably meant "products" rather "limits".
    $endgroup$
    – Arnaud D.
    yesterday










  • $begingroup$
    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
    yesterday






  • 1




    $begingroup$
    @ArnaudD.: Additive functors don't preserve (possibly infinite) products either, though!
    $endgroup$
    – Eric Wofsey
    23 hours ago










  • $begingroup$
    @EricWofsey Indeed! Derek's answer makes me wonder if perhaps the author meant "representable functors"...
    $endgroup$
    – Arnaud D.
    23 hours ago















$begingroup$
"Additive functors preserve limits" is completely false; the author probably meant "products" rather "limits".
$endgroup$
– Arnaud D.
yesterday




$begingroup$
"Additive functors preserve limits" is completely false; the author probably meant "products" rather "limits".
$endgroup$
– Arnaud D.
yesterday












$begingroup$
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
yesterday




$begingroup$
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
yesterday




1




1




$begingroup$
@ArnaudD.: Additive functors don't preserve (possibly infinite) products either, though!
$endgroup$
– Eric Wofsey
23 hours ago




$begingroup$
@ArnaudD.: Additive functors don't preserve (possibly infinite) products either, though!
$endgroup$
– Eric Wofsey
23 hours ago












$begingroup$
@EricWofsey Indeed! Derek's answer makes me wonder if perhaps the author meant "representable functors"...
$endgroup$
– Arnaud D.
23 hours ago




$begingroup$
@EricWofsey Indeed! Derek's answer makes me wonder if perhaps the author meant "representable functors"...
$endgroup$
– Arnaud D.
23 hours ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

If you know that the functor $T=operatornameHom_R(-,M):RmathttMod^opto mathttAb$ preserves limits, then the result follows. Indeed, the direct sum $bigoplus M_i$ is just the coproduct of the $M_i$ in $RmathttMod$ and thus the product of the $M_i$ in $RmathttMod^op$, 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 $RtextMod$ 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 $$mathsfHom^op(mathsfHom_R(N,P),M)=mathsfHom(M,mathsfHom_R(N,P))congmathsfHom(N,mathsfHom_R(M,P))$$ (natural in $M$ and $N$ [and $P$]) meaning $mathsfHom_R(-,P)$ 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
      23 hours ago










    • $begingroup$
      @EricWofsey You're right. I've added commutativity of $R$ as an assumption for now.
      $endgroup$
      – Derek Elkins
      23 hours ago











    Your Answer





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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

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    active

    oldest

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    4












    $begingroup$

    If you know that the functor $T=operatornameHom_R(-,M):RmathttMod^opto mathttAb$ preserves limits, then the result follows. Indeed, the direct sum $bigoplus M_i$ is just the coproduct of the $M_i$ in $RmathttMod$ and thus the product of the $M_i$ in $RmathttMod^op$, 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









    $endgroup$

















      4












      $begingroup$

      If you know that the functor $T=operatornameHom_R(-,M):RmathttMod^opto mathttAb$ preserves limits, then the result follows. Indeed, the direct sum $bigoplus M_i$ is just the coproduct of the $M_i$ in $RmathttMod$ and thus the product of the $M_i$ in $RmathttMod^op$, 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









      $endgroup$















        4












        4








        4





        $begingroup$

        If you know that the functor $T=operatornameHom_R(-,M):RmathttMod^opto mathttAb$ preserves limits, then the result follows. Indeed, the direct sum $bigoplus M_i$ is just the coproduct of the $M_i$ in $RmathttMod$ and thus the product of the $M_i$ in $RmathttMod^op$, 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 $T=operatornameHom_R(-,M):RmathttMod^opto mathttAb$ preserves limits, then the result follows. Indeed, the direct sum $bigoplus M_i$ is just the coproduct of the $M_i$ in $RmathttMod$ and thus the product of the $M_i$ in $RmathttMod^op$, 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 23 hours ago









        Eric WofseyEric Wofsey

        192k14217351




        192k14217351





















            3












            $begingroup$

            Here's a purely categorical approach when $R$ is commutative. (The proof below actually works for any symmetric monoidally closed category which $RtextMod$ 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 $$mathsfHom^op(mathsfHom_R(N,P),M)=mathsfHom(M,mathsfHom_R(N,P))congmathsfHom(N,mathsfHom_R(M,P))$$ (natural in $M$ and $N$ [and $P$]) meaning $mathsfHom_R(-,P)$ 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
              23 hours ago










            • $begingroup$
              @EricWofsey You're right. I've added commutativity of $R$ as an assumption for now.
              $endgroup$
              – Derek Elkins
              23 hours ago















            3












            $begingroup$

            Here's a purely categorical approach when $R$ is commutative. (The proof below actually works for any symmetric monoidally closed category which $RtextMod$ 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 $$mathsfHom^op(mathsfHom_R(N,P),M)=mathsfHom(M,mathsfHom_R(N,P))congmathsfHom(N,mathsfHom_R(M,P))$$ (natural in $M$ and $N$ [and $P$]) meaning $mathsfHom_R(-,P)$ 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
              23 hours ago










            • $begingroup$
              @EricWofsey You're right. I've added commutativity of $R$ as an assumption for now.
              $endgroup$
              – Derek Elkins
              23 hours ago













            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 $RtextMod$ 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 $$mathsfHom^op(mathsfHom_R(N,P),M)=mathsfHom(M,mathsfHom_R(N,P))congmathsfHom(N,mathsfHom_R(M,P))$$ (natural in $M$ and $N$ [and $P$]) meaning $mathsfHom_R(-,P)$ 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 $RtextMod$ 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 $$mathsfHom^op(mathsfHom_R(N,P),M)=mathsfHom(M,mathsfHom_R(N,P))congmathsfHom(N,mathsfHom_R(M,P))$$ (natural in $M$ and $N$ [and $P$]) meaning $mathsfHom_R(-,P)$ 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 23 hours ago

























            answered 23 hours ago









            Derek ElkinsDerek Elkins

            17.5k11437




            17.5k11437







            • 2




              $begingroup$
              $R$ seems to not be assumed to be commutative though, so this is not actually an internal Hom.
              $endgroup$
              – Eric Wofsey
              23 hours ago










            • $begingroup$
              @EricWofsey You're right. I've added commutativity of $R$ as an assumption for now.
              $endgroup$
              – Derek Elkins
              23 hours ago












            • 2




              $begingroup$
              $R$ seems to not be assumed to be commutative though, so this is not actually an internal Hom.
              $endgroup$
              – Eric Wofsey
              23 hours ago










            • $begingroup$
              @EricWofsey You're right. I've added commutativity of $R$ as an assumption for now.
              $endgroup$
              – Derek Elkins
              23 hours ago







            2




            2




            $begingroup$
            $R$ seems to not be assumed to be commutative though, so this is not actually an internal Hom.
            $endgroup$
            – Eric Wofsey
            23 hours ago




            $begingroup$
            $R$ seems to not be assumed to be commutative though, so this is not actually an internal Hom.
            $endgroup$
            – Eric Wofsey
            23 hours ago












            $begingroup$
            @EricWofsey You're right. I've added commutativity of $R$ as an assumption for now.
            $endgroup$
            – Derek Elkins
            23 hours ago




            $begingroup$
            @EricWofsey You're right. I've added commutativity of $R$ as an assumption for now.
            $endgroup$
            – Derek Elkins
            23 hours ago

















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