Understanding sheaves on a $2$-element set












4












$begingroup$


I'm working through the Geometry of Schemes and wanted some clarification for an exercise.



Exercise I-5 considers a two-element set $X={0,1}$ with the discrete topology and asks the reader to find the relations between the objects of a sheaf (of abelian groups) on $X$.



If we let $mathcal{F}$ be the sheaf, then it's clear that $mathcal{F}(emptyset)$ is the trivial group and that we have a commutative diagram of restrictions from $mathcal{F}({0})$ to $mathcal{F}(emptyset)$, from $mathcal{F}({1})$ to $mathcal{F}(emptyset)$, from $mathcal{F}({0,1})$ to $mathcal{F}({0})$, and from $mathcal{F}({0,1})$ to $mathcal{F}(emptyset)$.



By the sheaf axiom, it seems like for any $sinmathcal{F}({0})$ and $tinmathcal{F}({{1}})$, $s$ and $t$ restricted to the intersection, $emptyset$, must be the same, so there is some unique section in $mathcal{F}({0,1})$ that restricts to $s$ and $t$.



What does this say about $mathcal{F}({0,1})$? I'm guessing it's related to the fiber product, but I'm not particularly well-versed in category theory. Also, how does this generalize to sheaves over different categories?






algebraic-geometry category-theory sheaf-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you haven't already, it is easy to prove that $mathcal F(varnothing)$ is always the terminal object for any topological space $X$ for sheaves of sets. You can also prove that if this is true for sheaves of sets, then it's also true for sheaves of any essentially algebraic structure.
    $endgroup$
    – Derek Elkins
    1 hour ago
















4












$begingroup$


I'm working through the Geometry of Schemes and wanted some clarification for an exercise.



Exercise I-5 considers a two-element set $X={0,1}$ with the discrete topology and asks the reader to find the relations between the objects of a sheaf (of abelian groups) on $X$.



If we let $mathcal{F}$ be the sheaf, then it's clear that $mathcal{F}(emptyset)$ is the trivial group and that we have a commutative diagram of restrictions from $mathcal{F}({0})$ to $mathcal{F}(emptyset)$, from $mathcal{F}({1})$ to $mathcal{F}(emptyset)$, from $mathcal{F}({0,1})$ to $mathcal{F}({0})$, and from $mathcal{F}({0,1})$ to $mathcal{F}(emptyset)$.



By the sheaf axiom, it seems like for any $sinmathcal{F}({0})$ and $tinmathcal{F}({{1}})$, $s$ and $t$ restricted to the intersection, $emptyset$, must be the same, so there is some unique section in $mathcal{F}({0,1})$ that restricts to $s$ and $t$.



What does this say about $mathcal{F}({0,1})$? I'm guessing it's related to the fiber product, but I'm not particularly well-versed in category theory. Also, how does this generalize to sheaves over different categories?






algebraic-geometry category-theory sheaf-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you haven't already, it is easy to prove that $mathcal F(varnothing)$ is always the terminal object for any topological space $X$ for sheaves of sets. You can also prove that if this is true for sheaves of sets, then it's also true for sheaves of any essentially algebraic structure.
    $endgroup$
    – Derek Elkins
    1 hour ago














4












4








4





$begingroup$


I'm working through the Geometry of Schemes and wanted some clarification for an exercise.



Exercise I-5 considers a two-element set $X={0,1}$ with the discrete topology and asks the reader to find the relations between the objects of a sheaf (of abelian groups) on $X$.



If we let $mathcal{F}$ be the sheaf, then it's clear that $mathcal{F}(emptyset)$ is the trivial group and that we have a commutative diagram of restrictions from $mathcal{F}({0})$ to $mathcal{F}(emptyset)$, from $mathcal{F}({1})$ to $mathcal{F}(emptyset)$, from $mathcal{F}({0,1})$ to $mathcal{F}({0})$, and from $mathcal{F}({0,1})$ to $mathcal{F}(emptyset)$.



By the sheaf axiom, it seems like for any $sinmathcal{F}({0})$ and $tinmathcal{F}({{1}})$, $s$ and $t$ restricted to the intersection, $emptyset$, must be the same, so there is some unique section in $mathcal{F}({0,1})$ that restricts to $s$ and $t$.



What does this say about $mathcal{F}({0,1})$? I'm guessing it's related to the fiber product, but I'm not particularly well-versed in category theory. Also, how does this generalize to sheaves over different categories?






algebraic-geometry category-theory sheaf-theory







share|cite|improve this question











$endgroup$




I'm working through the Geometry of Schemes and wanted some clarification for an exercise.



Exercise I-5 considers a two-element set $X={0,1}$ with the discrete topology and asks the reader to find the relations between the objects of a sheaf (of abelian groups) on $X$.



If we let $mathcal{F}$ be the sheaf, then it's clear that $mathcal{F}(emptyset)$ is the trivial group and that we have a commutative diagram of restrictions from $mathcal{F}({0})$ to $mathcal{F}(emptyset)$, from $mathcal{F}({1})$ to $mathcal{F}(emptyset)$, from $mathcal{F}({0,1})$ to $mathcal{F}({0})$, and from $mathcal{F}({0,1})$ to $mathcal{F}(emptyset)$.



By the sheaf axiom, it seems like for any $sinmathcal{F}({0})$ and $tinmathcal{F}({{1}})$, $s$ and $t$ restricted to the intersection, $emptyset$, must be the same, so there is some unique section in $mathcal{F}({0,1})$ that restricts to $s$ and $t$.



What does this say about $mathcal{F}({0,1})$? I'm guessing it's related to the fiber product, but I'm not particularly well-versed in category theory. Also, how does this generalize to sheaves over different categories?






algebraic-geometry category-theory sheaf-theory




algebraic-geometry category-theory sheaf-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 4 hours ago









Asaf Karagila♦

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303k32429762










asked 6 hours ago









whethamwhetham

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  • $begingroup$
    If you haven't already, it is easy to prove that $mathcal F(varnothing)$ is always the terminal object for any topological space $X$ for sheaves of sets. You can also prove that if this is true for sheaves of sets, then it's also true for sheaves of any essentially algebraic structure.
    $endgroup$
    – Derek Elkins
    1 hour ago


















  • $begingroup$
    If you haven't already, it is easy to prove that $mathcal F(varnothing)$ is always the terminal object for any topological space $X$ for sheaves of sets. You can also prove that if this is true for sheaves of sets, then it's also true for sheaves of any essentially algebraic structure.
    $endgroup$
    – Derek Elkins
    1 hour ago
















$begingroup$
If you haven't already, it is easy to prove that $mathcal F(varnothing)$ is always the terminal object for any topological space $X$ for sheaves of sets. You can also prove that if this is true for sheaves of sets, then it's also true for sheaves of any essentially algebraic structure.
$endgroup$
– Derek Elkins
1 hour ago




$begingroup$
If you haven't already, it is easy to prove that $mathcal F(varnothing)$ is always the terminal object for any topological space $X$ for sheaves of sets. You can also prove that if this is true for sheaves of sets, then it's also true for sheaves of any essentially algebraic structure.
$endgroup$
– Derek Elkins
1 hour ago










1 Answer
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oldest

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5












$begingroup$

In this case, $mathcal{F}({0,1})$ is just the direct product of the groups
$mathcal{F}({0})$ and $mathcal{F}({1})$.



This commutative diagram has to be a pullback
$require{AMScd}$
begin{CD}
mathcal{F}({0,1}) @>>> mathcal{F}({1})\
@V V V @VV V\
mathcal{F}({0}) @>>> mathcal{F}(emptyset)={0}
end{CD}
but as the southeast corner is trivial, the northwest group
is the direct product of the other corners, and the nontrivial
maps are the projections from a direct product to its factors.






share|cite|improve this answer









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    1 Answer
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    5












    $begingroup$

    In this case, $mathcal{F}({0,1})$ is just the direct product of the groups
    $mathcal{F}({0})$ and $mathcal{F}({1})$.



    This commutative diagram has to be a pullback
    $require{AMScd}$
    begin{CD}
    mathcal{F}({0,1}) @>>> mathcal{F}({1})\
    @V V V @VV V\
    mathcal{F}({0}) @>>> mathcal{F}(emptyset)={0}
    end{CD}
    but as the southeast corner is trivial, the northwest group
    is the direct product of the other corners, and the nontrivial
    maps are the projections from a direct product to its factors.






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      In this case, $mathcal{F}({0,1})$ is just the direct product of the groups
      $mathcal{F}({0})$ and $mathcal{F}({1})$.



      This commutative diagram has to be a pullback
      $require{AMScd}$
      begin{CD}
      mathcal{F}({0,1}) @>>> mathcal{F}({1})\
      @V V V @VV V\
      mathcal{F}({0}) @>>> mathcal{F}(emptyset)={0}
      end{CD}
      but as the southeast corner is trivial, the northwest group
      is the direct product of the other corners, and the nontrivial
      maps are the projections from a direct product to its factors.






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        In this case, $mathcal{F}({0,1})$ is just the direct product of the groups
        $mathcal{F}({0})$ and $mathcal{F}({1})$.



        This commutative diagram has to be a pullback
        $require{AMScd}$
        begin{CD}
        mathcal{F}({0,1}) @>>> mathcal{F}({1})\
        @V V V @VV V\
        mathcal{F}({0}) @>>> mathcal{F}(emptyset)={0}
        end{CD}
        but as the southeast corner is trivial, the northwest group
        is the direct product of the other corners, and the nontrivial
        maps are the projections from a direct product to its factors.






        share|cite|improve this answer









        $endgroup$



        In this case, $mathcal{F}({0,1})$ is just the direct product of the groups
        $mathcal{F}({0})$ and $mathcal{F}({1})$.



        This commutative diagram has to be a pullback
        $require{AMScd}$
        begin{CD}
        mathcal{F}({0,1}) @>>> mathcal{F}({1})\
        @V V V @VV V\
        mathcal{F}({0}) @>>> mathcal{F}(emptyset)={0}
        end{CD}
        but as the southeast corner is trivial, the northwest group
        is the direct product of the other corners, and the nontrivial
        maps are the projections from a direct product to its factors.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 5 hours ago









        Lord Shark the UnknownLord Shark the Unknown

        104k1160132




        104k1160132






























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