Exercise on monomorphism of sheaves

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The following is exercise 6, chapter 0 from Johnstone's Topos Theory:




Suppose $Frightarrowtail G$ a monomorphism of sheaves in $Shv(X)$ [$X$ being a topological space], and $sigmain G(U)$ for some open $Usubseteq X$. Prove that there is a unique largest open set $Vsubseteq U$ such that $sigma_V$ is in the image of $F(V)$.




Unicity is easy, because given a family of such subsets $V_i_iin I$ then the union $bigcup_I V_i$ would share the same property. However, I can't find a reason for the existence to hold. I tried with the following counterexample:



$X=*$ the singleton space



$Fin Shv(X)$ the constant sheaf sending all opens to the set $a$



$Gin Shv(X)$ the constant sheaf sending all opens to $a,sigma$



Then $Frightarrowtail G$, but a $V$ as above does not exist for any choice of $Usubseteq X$ and taking $sigmain G(U)$.







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    The following is exercise 6, chapter 0 from Johnstone's Topos Theory:




    Suppose $Frightarrowtail G$ a monomorphism of sheaves in $Shv(X)$ [$X$ being a topological space], and $sigmain G(U)$ for some open $Usubseteq X$. Prove that there is a unique largest open set $Vsubseteq U$ such that $sigma_V$ is in the image of $F(V)$.




    Unicity is easy, because given a family of such subsets $V_i_iin I$ then the union $bigcup_I V_i$ would share the same property. However, I can't find a reason for the existence to hold. I tried with the following counterexample:



    $X=*$ the singleton space



    $Fin Shv(X)$ the constant sheaf sending all opens to the set $a$



    $Gin Shv(X)$ the constant sheaf sending all opens to $a,sigma$



    Then $Frightarrowtail G$, but a $V$ as above does not exist for any choice of $Usubseteq X$ and taking $sigmain G(U)$.







    share|cite|improve this question
























      up vote
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      down vote

      favorite









      up vote
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      down vote

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      The following is exercise 6, chapter 0 from Johnstone's Topos Theory:




      Suppose $Frightarrowtail G$ a monomorphism of sheaves in $Shv(X)$ [$X$ being a topological space], and $sigmain G(U)$ for some open $Usubseteq X$. Prove that there is a unique largest open set $Vsubseteq U$ such that $sigma_V$ is in the image of $F(V)$.




      Unicity is easy, because given a family of such subsets $V_i_iin I$ then the union $bigcup_I V_i$ would share the same property. However, I can't find a reason for the existence to hold. I tried with the following counterexample:



      $X=*$ the singleton space



      $Fin Shv(X)$ the constant sheaf sending all opens to the set $a$



      $Gin Shv(X)$ the constant sheaf sending all opens to $a,sigma$



      Then $Frightarrowtail G$, but a $V$ as above does not exist for any choice of $Usubseteq X$ and taking $sigmain G(U)$.







      share|cite|improve this question














      The following is exercise 6, chapter 0 from Johnstone's Topos Theory:




      Suppose $Frightarrowtail G$ a monomorphism of sheaves in $Shv(X)$ [$X$ being a topological space], and $sigmain G(U)$ for some open $Usubseteq X$. Prove that there is a unique largest open set $Vsubseteq U$ such that $sigma_V$ is in the image of $F(V)$.




      Unicity is easy, because given a family of such subsets $V_i_iin I$ then the union $bigcup_I V_i$ would share the same property. However, I can't find a reason for the existence to hold. I tried with the following counterexample:



      $X=*$ the singleton space



      $Fin Shv(X)$ the constant sheaf sending all opens to the set $a$



      $Gin Shv(X)$ the constant sheaf sending all opens to $a,sigma$



      Then $Frightarrowtail G$, but a $V$ as above does not exist for any choice of $Usubseteq X$ and taking $sigmain G(U)$.









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 16 at 10:06

























      asked Aug 16 at 9:38









      TheMadcapLaughs

      341211




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          If $A$ is a sheaf, then $A(varnothing)$ is a one-point set.



          So your counterexample is flawed; you give a formula for a constant presheaf, but it is not a sheaf. You have to take the sheafification to get a constant sheaf; e.g. $G(*) = a, sigma $ and $G(varnothing) = * $.



          Going back to the generic problem, $sigma|_varnothing$ is the unique element of $G(varnothing)$, which is the image of the unique element of $F(varnothing)$, so there's your existence proof.






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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote



            accepted










            If $A$ is a sheaf, then $A(varnothing)$ is a one-point set.



            So your counterexample is flawed; you give a formula for a constant presheaf, but it is not a sheaf. You have to take the sheafification to get a constant sheaf; e.g. $G(*) = a, sigma $ and $G(varnothing) = * $.



            Going back to the generic problem, $sigma|_varnothing$ is the unique element of $G(varnothing)$, which is the image of the unique element of $F(varnothing)$, so there's your existence proof.






            share|cite|improve this answer
























              up vote
              1
              down vote



              accepted










              If $A$ is a sheaf, then $A(varnothing)$ is a one-point set.



              So your counterexample is flawed; you give a formula for a constant presheaf, but it is not a sheaf. You have to take the sheafification to get a constant sheaf; e.g. $G(*) = a, sigma $ and $G(varnothing) = * $.



              Going back to the generic problem, $sigma|_varnothing$ is the unique element of $G(varnothing)$, which is the image of the unique element of $F(varnothing)$, so there's your existence proof.






              share|cite|improve this answer






















                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                If $A$ is a sheaf, then $A(varnothing)$ is a one-point set.



                So your counterexample is flawed; you give a formula for a constant presheaf, but it is not a sheaf. You have to take the sheafification to get a constant sheaf; e.g. $G(*) = a, sigma $ and $G(varnothing) = * $.



                Going back to the generic problem, $sigma|_varnothing$ is the unique element of $G(varnothing)$, which is the image of the unique element of $F(varnothing)$, so there's your existence proof.






                share|cite|improve this answer












                If $A$ is a sheaf, then $A(varnothing)$ is a one-point set.



                So your counterexample is flawed; you give a formula for a constant presheaf, but it is not a sheaf. You have to take the sheafification to get a constant sheaf; e.g. $G(*) = a, sigma $ and $G(varnothing) = * $.



                Going back to the generic problem, $sigma|_varnothing$ is the unique element of $G(varnothing)$, which is the image of the unique element of $F(varnothing)$, so there's your existence proof.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 16 at 9:46









                Hurkyl

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