Left and right adjoint of the presheaf evaluation $widehatC to mathbfSet$, $F mapsto F(c)$

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I would like to find the left and right adjoint of the presheaf evaluation functor $F colon widehatC rightarrow mathbfSet$, $X mapsto X(c)$ for a fixed object $c in operatornameob(C)$, where $widehatC := mathbfSet^C^mathrmop$.



I could not understand this proof.



To show $F$ is right adjoint we define $L_c colon mathbfSet rightarrow widehatC$ such that
$$
operatornameHom(L_c(X), D)
cong operatornameHom(X, F(D)
= operatornameHom(X,D(c)).
$$




The cited proof states that we define $L_c(X)= X cdot L_c(1) := bigsqcup_x in X L_c(1)$.



I don't know what this actually means—what is $L_c(1)$ anyways? Secondly, I don't see how this defines an element in $widehatC$. I am equally confused with the construction of right adjoint. Elaboration is really appreciated.










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    I would like to find the left and right adjoint of the presheaf evaluation functor $F colon widehatC rightarrow mathbfSet$, $X mapsto X(c)$ for a fixed object $c in operatornameob(C)$, where $widehatC := mathbfSet^C^mathrmop$.



    I could not understand this proof.



    To show $F$ is right adjoint we define $L_c colon mathbfSet rightarrow widehatC$ such that
    $$
    operatornameHom(L_c(X), D)
    cong operatornameHom(X, F(D)
    = operatornameHom(X,D(c)).
    $$




    The cited proof states that we define $L_c(X)= X cdot L_c(1) := bigsqcup_x in X L_c(1)$.



    I don't know what this actually means—what is $L_c(1)$ anyways? Secondly, I don't see how this defines an element in $widehatC$. I am equally confused with the construction of right adjoint. Elaboration is really appreciated.










    share|cite|improve this question

























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I would like to find the left and right adjoint of the presheaf evaluation functor $F colon widehatC rightarrow mathbfSet$, $X mapsto X(c)$ for a fixed object $c in operatornameob(C)$, where $widehatC := mathbfSet^C^mathrmop$.



      I could not understand this proof.



      To show $F$ is right adjoint we define $L_c colon mathbfSet rightarrow widehatC$ such that
      $$
      operatornameHom(L_c(X), D)
      cong operatornameHom(X, F(D)
      = operatornameHom(X,D(c)).
      $$




      The cited proof states that we define $L_c(X)= X cdot L_c(1) := bigsqcup_x in X L_c(1)$.



      I don't know what this actually means—what is $L_c(1)$ anyways? Secondly, I don't see how this defines an element in $widehatC$. I am equally confused with the construction of right adjoint. Elaboration is really appreciated.










      share|cite|improve this question















      I would like to find the left and right adjoint of the presheaf evaluation functor $F colon widehatC rightarrow mathbfSet$, $X mapsto X(c)$ for a fixed object $c in operatornameob(C)$, where $widehatC := mathbfSet^C^mathrmop$.



      I could not understand this proof.



      To show $F$ is right adjoint we define $L_c colon mathbfSet rightarrow widehatC$ such that
      $$
      operatornameHom(L_c(X), D)
      cong operatornameHom(X, F(D)
      = operatornameHom(X,D(c)).
      $$




      The cited proof states that we define $L_c(X)= X cdot L_c(1) := bigsqcup_x in X L_c(1)$.



      I don't know what this actually means—what is $L_c(1)$ anyways? Secondly, I don't see how this defines an element in $widehatC$. I am equally confused with the construction of right adjoint. Elaboration is really appreciated.







      category-theory sheaf-theory






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      edited Aug 31 at 12:19









      Jendrik Stelzner

      7,69121137




      7,69121137










      asked Aug 31 at 8:23









      Cyryl L.

      1,8072821




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          2 Answers
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          2
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          Any left adjoint preserves coproducts. Since any set $X$ is build up as a coproduct $X=bigsqcup_xin X 1$ of a bunch of copies of the singleton $1$ (i.e. the set containing precisely one element), it suffices to specify what value $L_c$ takes on $1$. The computation in the linked answer shows that we must have $L_c(1)=hom(-,c)$, which is clearly a presheaf on $C$. Thus, in summary, for any set $X$ we have $L_c(X)=bigsqcup_xin X hom(-,c)$, where the coproduct is taken in the presheaf category.






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            up vote
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            down vote













            To give a more precise answer, I think the easiest way to define the adjoint is to first find out what it has to be, and then check that the definition we have found indeed works.



            So suppose there is a left adjoint $L_c$ to the evaluation functor. Given that $L_c$ is a left adjoint, it has to preserve colimits, and both the category of sets and the presheaf category happen to have colimits. But we also know that any set $X$ can be obtained as a colimit $Xsimeq coprod_xin X1$, where $1$ is the singleton. So we necessarily have that $L_c(X) simeq coprod_xin XL_c(1)$ in the presheaf category $widehatC$. So it suffices to determine $L_c(1)$ to determine the entire functor.



            The adjunction $operatornameHom_widehatC(L_c(X),D) simeq operatornameHom_mathbfSet(X,D(c))$ specialised to $X = 1$ gives then $operatornameHom_widehatC(L_c(1),D)simeq operatornameHom_mathbfSet(1,D(c)) simeq D(c)$. This lets us guess what $L_c(1)$ should be as we happen to know an presheaf satisfying exactly this equation for any other presheaf $D$. This is given by the Yoneda lemma :



            If we denote $y: C to widehatC$ the Yoneda embedding defined by $y(c) = operatornameHom_C(-,c)$, then the Yoneda lemma states that $operatornameHom_widehatC(y(c),D)simeq D(c)$, which is exactly what we wanted.



            We can then reconstruct the entire $L_c$ from these observations, by taking $L_c(X)simeq coprod_xin Xy(c)$. Now we just have to check that our guess was the right one, and $L_c$ is indeed the left adjoint to the evaluation functor. This is given by the following isomorphisms:
            $$
            operatornameHom_widehat C
            left(
            coprod_xin X y(c), D
            right)
            simeq prod_xin X operatornameHom_widehatC(y(c),D)
            simeq prod_xin X D(c)
            simeq D(c)^X
            simeq operatornameHom(X,D(C))
            $$






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              2 Answers
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              2 Answers
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              active

              oldest

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              up vote
              2
              down vote













              Any left adjoint preserves coproducts. Since any set $X$ is build up as a coproduct $X=bigsqcup_xin X 1$ of a bunch of copies of the singleton $1$ (i.e. the set containing precisely one element), it suffices to specify what value $L_c$ takes on $1$. The computation in the linked answer shows that we must have $L_c(1)=hom(-,c)$, which is clearly a presheaf on $C$. Thus, in summary, for any set $X$ we have $L_c(X)=bigsqcup_xin X hom(-,c)$, where the coproduct is taken in the presheaf category.






              share|cite|improve this answer


























                up vote
                2
                down vote













                Any left adjoint preserves coproducts. Since any set $X$ is build up as a coproduct $X=bigsqcup_xin X 1$ of a bunch of copies of the singleton $1$ (i.e. the set containing precisely one element), it suffices to specify what value $L_c$ takes on $1$. The computation in the linked answer shows that we must have $L_c(1)=hom(-,c)$, which is clearly a presheaf on $C$. Thus, in summary, for any set $X$ we have $L_c(X)=bigsqcup_xin X hom(-,c)$, where the coproduct is taken in the presheaf category.






                share|cite|improve this answer
























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  Any left adjoint preserves coproducts. Since any set $X$ is build up as a coproduct $X=bigsqcup_xin X 1$ of a bunch of copies of the singleton $1$ (i.e. the set containing precisely one element), it suffices to specify what value $L_c$ takes on $1$. The computation in the linked answer shows that we must have $L_c(1)=hom(-,c)$, which is clearly a presheaf on $C$. Thus, in summary, for any set $X$ we have $L_c(X)=bigsqcup_xin X hom(-,c)$, where the coproduct is taken in the presheaf category.






                  share|cite|improve this answer














                  Any left adjoint preserves coproducts. Since any set $X$ is build up as a coproduct $X=bigsqcup_xin X 1$ of a bunch of copies of the singleton $1$ (i.e. the set containing precisely one element), it suffices to specify what value $L_c$ takes on $1$. The computation in the linked answer shows that we must have $L_c(1)=hom(-,c)$, which is clearly a presheaf on $C$. Thus, in summary, for any set $X$ we have $L_c(X)=bigsqcup_xin X hom(-,c)$, where the coproduct is taken in the presheaf category.







                  share|cite|improve this answer














                  share|cite|improve this answer



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                  edited Aug 31 at 8:54

























                  answered Aug 31 at 8:49









                  asdq

                  1,5691417




                  1,5691417




















                      up vote
                      1
                      down vote













                      To give a more precise answer, I think the easiest way to define the adjoint is to first find out what it has to be, and then check that the definition we have found indeed works.



                      So suppose there is a left adjoint $L_c$ to the evaluation functor. Given that $L_c$ is a left adjoint, it has to preserve colimits, and both the category of sets and the presheaf category happen to have colimits. But we also know that any set $X$ can be obtained as a colimit $Xsimeq coprod_xin X1$, where $1$ is the singleton. So we necessarily have that $L_c(X) simeq coprod_xin XL_c(1)$ in the presheaf category $widehatC$. So it suffices to determine $L_c(1)$ to determine the entire functor.



                      The adjunction $operatornameHom_widehatC(L_c(X),D) simeq operatornameHom_mathbfSet(X,D(c))$ specialised to $X = 1$ gives then $operatornameHom_widehatC(L_c(1),D)simeq operatornameHom_mathbfSet(1,D(c)) simeq D(c)$. This lets us guess what $L_c(1)$ should be as we happen to know an presheaf satisfying exactly this equation for any other presheaf $D$. This is given by the Yoneda lemma :



                      If we denote $y: C to widehatC$ the Yoneda embedding defined by $y(c) = operatornameHom_C(-,c)$, then the Yoneda lemma states that $operatornameHom_widehatC(y(c),D)simeq D(c)$, which is exactly what we wanted.



                      We can then reconstruct the entire $L_c$ from these observations, by taking $L_c(X)simeq coprod_xin Xy(c)$. Now we just have to check that our guess was the right one, and $L_c$ is indeed the left adjoint to the evaluation functor. This is given by the following isomorphisms:
                      $$
                      operatornameHom_widehat C
                      left(
                      coprod_xin X y(c), D
                      right)
                      simeq prod_xin X operatornameHom_widehatC(y(c),D)
                      simeq prod_xin X D(c)
                      simeq D(c)^X
                      simeq operatornameHom(X,D(C))
                      $$






                      share|cite|improve this answer


























                        up vote
                        1
                        down vote













                        To give a more precise answer, I think the easiest way to define the adjoint is to first find out what it has to be, and then check that the definition we have found indeed works.



                        So suppose there is a left adjoint $L_c$ to the evaluation functor. Given that $L_c$ is a left adjoint, it has to preserve colimits, and both the category of sets and the presheaf category happen to have colimits. But we also know that any set $X$ can be obtained as a colimit $Xsimeq coprod_xin X1$, where $1$ is the singleton. So we necessarily have that $L_c(X) simeq coprod_xin XL_c(1)$ in the presheaf category $widehatC$. So it suffices to determine $L_c(1)$ to determine the entire functor.



                        The adjunction $operatornameHom_widehatC(L_c(X),D) simeq operatornameHom_mathbfSet(X,D(c))$ specialised to $X = 1$ gives then $operatornameHom_widehatC(L_c(1),D)simeq operatornameHom_mathbfSet(1,D(c)) simeq D(c)$. This lets us guess what $L_c(1)$ should be as we happen to know an presheaf satisfying exactly this equation for any other presheaf $D$. This is given by the Yoneda lemma :



                        If we denote $y: C to widehatC$ the Yoneda embedding defined by $y(c) = operatornameHom_C(-,c)$, then the Yoneda lemma states that $operatornameHom_widehatC(y(c),D)simeq D(c)$, which is exactly what we wanted.



                        We can then reconstruct the entire $L_c$ from these observations, by taking $L_c(X)simeq coprod_xin Xy(c)$. Now we just have to check that our guess was the right one, and $L_c$ is indeed the left adjoint to the evaluation functor. This is given by the following isomorphisms:
                        $$
                        operatornameHom_widehat C
                        left(
                        coprod_xin X y(c), D
                        right)
                        simeq prod_xin X operatornameHom_widehatC(y(c),D)
                        simeq prod_xin X D(c)
                        simeq D(c)^X
                        simeq operatornameHom(X,D(C))
                        $$






                        share|cite|improve this answer
























                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          To give a more precise answer, I think the easiest way to define the adjoint is to first find out what it has to be, and then check that the definition we have found indeed works.



                          So suppose there is a left adjoint $L_c$ to the evaluation functor. Given that $L_c$ is a left adjoint, it has to preserve colimits, and both the category of sets and the presheaf category happen to have colimits. But we also know that any set $X$ can be obtained as a colimit $Xsimeq coprod_xin X1$, where $1$ is the singleton. So we necessarily have that $L_c(X) simeq coprod_xin XL_c(1)$ in the presheaf category $widehatC$. So it suffices to determine $L_c(1)$ to determine the entire functor.



                          The adjunction $operatornameHom_widehatC(L_c(X),D) simeq operatornameHom_mathbfSet(X,D(c))$ specialised to $X = 1$ gives then $operatornameHom_widehatC(L_c(1),D)simeq operatornameHom_mathbfSet(1,D(c)) simeq D(c)$. This lets us guess what $L_c(1)$ should be as we happen to know an presheaf satisfying exactly this equation for any other presheaf $D$. This is given by the Yoneda lemma :



                          If we denote $y: C to widehatC$ the Yoneda embedding defined by $y(c) = operatornameHom_C(-,c)$, then the Yoneda lemma states that $operatornameHom_widehatC(y(c),D)simeq D(c)$, which is exactly what we wanted.



                          We can then reconstruct the entire $L_c$ from these observations, by taking $L_c(X)simeq coprod_xin Xy(c)$. Now we just have to check that our guess was the right one, and $L_c$ is indeed the left adjoint to the evaluation functor. This is given by the following isomorphisms:
                          $$
                          operatornameHom_widehat C
                          left(
                          coprod_xin X y(c), D
                          right)
                          simeq prod_xin X operatornameHom_widehatC(y(c),D)
                          simeq prod_xin X D(c)
                          simeq D(c)^X
                          simeq operatornameHom(X,D(C))
                          $$






                          share|cite|improve this answer














                          To give a more precise answer, I think the easiest way to define the adjoint is to first find out what it has to be, and then check that the definition we have found indeed works.



                          So suppose there is a left adjoint $L_c$ to the evaluation functor. Given that $L_c$ is a left adjoint, it has to preserve colimits, and both the category of sets and the presheaf category happen to have colimits. But we also know that any set $X$ can be obtained as a colimit $Xsimeq coprod_xin X1$, where $1$ is the singleton. So we necessarily have that $L_c(X) simeq coprod_xin XL_c(1)$ in the presheaf category $widehatC$. So it suffices to determine $L_c(1)$ to determine the entire functor.



                          The adjunction $operatornameHom_widehatC(L_c(X),D) simeq operatornameHom_mathbfSet(X,D(c))$ specialised to $X = 1$ gives then $operatornameHom_widehatC(L_c(1),D)simeq operatornameHom_mathbfSet(1,D(c)) simeq D(c)$. This lets us guess what $L_c(1)$ should be as we happen to know an presheaf satisfying exactly this equation for any other presheaf $D$. This is given by the Yoneda lemma :



                          If we denote $y: C to widehatC$ the Yoneda embedding defined by $y(c) = operatornameHom_C(-,c)$, then the Yoneda lemma states that $operatornameHom_widehatC(y(c),D)simeq D(c)$, which is exactly what we wanted.



                          We can then reconstruct the entire $L_c$ from these observations, by taking $L_c(X)simeq coprod_xin Xy(c)$. Now we just have to check that our guess was the right one, and $L_c$ is indeed the left adjoint to the evaluation functor. This is given by the following isomorphisms:
                          $$
                          operatornameHom_widehat C
                          left(
                          coprod_xin X y(c), D
                          right)
                          simeq prod_xin X operatornameHom_widehatC(y(c),D)
                          simeq prod_xin X D(c)
                          simeq D(c)^X
                          simeq operatornameHom(X,D(C))
                          $$







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                          edited Aug 31 at 14:10

























                          answered Aug 31 at 12:02









                          Thibaut Benjamin

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