If restricted morphism of ringed spaces are equal, then they are actually equal












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Given any two ringed spaces $(X, mathcal O_X)$ and $(Y, mathcal O_Y)$, let ${ U_lambda}_{lambda in Lambda}$ be an open covering of the topological space $X$. If $$f,g: (X, mathcal O_X) to (Y, mathcal O_Y)$$ are morphism of ringed spaces such that $$f|_{lambda} = g|_{lambda}, forall lambda in Lambda,$$ then prove that $$f=g.$$



I'm stuck after writing definitions in detail. Any help or hint will be appreciated.






sheaf-theory ringed-spaces







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    $begingroup$


    Given any two ringed spaces $(X, mathcal O_X)$ and $(Y, mathcal O_Y)$, let ${ U_lambda}_{lambda in Lambda}$ be an open covering of the topological space $X$. If $$f,g: (X, mathcal O_X) to (Y, mathcal O_Y)$$ are morphism of ringed spaces such that $$f|_{lambda} = g|_{lambda}, forall lambda in Lambda,$$ then prove that $$f=g.$$



    I'm stuck after writing definitions in detail. Any help or hint will be appreciated.






    sheaf-theory ringed-spaces







    share|cite|improve this question











    $endgroup$















      1












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      1





      $begingroup$


      Given any two ringed spaces $(X, mathcal O_X)$ and $(Y, mathcal O_Y)$, let ${ U_lambda}_{lambda in Lambda}$ be an open covering of the topological space $X$. If $$f,g: (X, mathcal O_X) to (Y, mathcal O_Y)$$ are morphism of ringed spaces such that $$f|_{lambda} = g|_{lambda}, forall lambda in Lambda,$$ then prove that $$f=g.$$



      I'm stuck after writing definitions in detail. Any help or hint will be appreciated.






      sheaf-theory ringed-spaces







      share|cite|improve this question











      $endgroup$




      Given any two ringed spaces $(X, mathcal O_X)$ and $(Y, mathcal O_Y)$, let ${ U_lambda}_{lambda in Lambda}$ be an open covering of the topological space $X$. If $$f,g: (X, mathcal O_X) to (Y, mathcal O_Y)$$ are morphism of ringed spaces such that $$f|_{lambda} = g|_{lambda}, forall lambda in Lambda,$$ then prove that $$f=g.$$



      I'm stuck after writing definitions in detail. Any help or hint will be appreciated.






      sheaf-theory ringed-spaces




      sheaf-theory ringed-spaces






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      share|cite|improve this question













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      edited Nov 30 '18 at 18:53









      Armando j18eos

      2,63511328




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      asked Nov 30 '18 at 18:00







      user621469





























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          $begingroup$

          First, I should hope its clear that if two maps of topological spaces agree on an open cover then they are equal (since maps of topological spaces are functions, so if they agree on any cover of their domain, they are equal as functions).



          Thus the key part is to show that $f^sharp=g^sharp$ as morphisms from $newcommandcalO{mathcal{O}}calO_Yto f_*calO_X=g_*calO_X$.



          Let $V$ be an open subset of $Y$. Let $aincalO_Y(V)$. Let $newcommandinv{^{-1}}U=finv(V)=ginv(V)$. Then $f^sharp(a),g^sharp(a)in calO_X(U)$ by definition. Then ${U_lambdacap U}_{lambdainLambda}$ gives an open cover of $U$, and since $f|_{U_lambda}=g|_{U_lambda}$, we have that $f^sharp(a)|_{Ucap U_lambda}=g^sharp(a)|_{Ucap U_lambda}$. Then since $f^sharp(a)$ and $g^sharp(a)$ are equal on an open cover of $U$, we must have that $f^sharp(a)=g^sharp(a)$ since $calO_X$ is a sheaf. Thus since $a$ and $V$ were arbitrary, $f^sharp=g^sharp$.



          Edit in response to comment:



          To clarify why $f^sharp(a)$ agreeing with $g^sharp(a)$ on an open cover of $U$ implies that they are equal, this is one of the axioms of sheaves. On wiki, this is the locality axiom. The axiom says that if $F$ is a sheaf, and $a,bin F(U)$, and ${U_i}$ is a cover of $U$, then if $a|_{U_i}=b|_{U_i}$ for all $i$, then $a=b$.






          share|cite|improve this answer











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          • $begingroup$
            Why "since $f^sharp(a)$ and $g^sharp(a)$ are equal on an open cover of $U$, we must have that $f^sharp(a)=g^sharp(a)$ since $calO_X$ is a sheaf" holds automatic? I couldn't say this impliance from the definition of the sheaf.
            $endgroup$
            – user621469
            Dec 1 '18 at 22:45











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          $begingroup$

          First, I should hope its clear that if two maps of topological spaces agree on an open cover then they are equal (since maps of topological spaces are functions, so if they agree on any cover of their domain, they are equal as functions).



          Thus the key part is to show that $f^sharp=g^sharp$ as morphisms from $newcommandcalO{mathcal{O}}calO_Yto f_*calO_X=g_*calO_X$.



          Let $V$ be an open subset of $Y$. Let $aincalO_Y(V)$. Let $newcommandinv{^{-1}}U=finv(V)=ginv(V)$. Then $f^sharp(a),g^sharp(a)in calO_X(U)$ by definition. Then ${U_lambdacap U}_{lambdainLambda}$ gives an open cover of $U$, and since $f|_{U_lambda}=g|_{U_lambda}$, we have that $f^sharp(a)|_{Ucap U_lambda}=g^sharp(a)|_{Ucap U_lambda}$. Then since $f^sharp(a)$ and $g^sharp(a)$ are equal on an open cover of $U$, we must have that $f^sharp(a)=g^sharp(a)$ since $calO_X$ is a sheaf. Thus since $a$ and $V$ were arbitrary, $f^sharp=g^sharp$.



          Edit in response to comment:



          To clarify why $f^sharp(a)$ agreeing with $g^sharp(a)$ on an open cover of $U$ implies that they are equal, this is one of the axioms of sheaves. On wiki, this is the locality axiom. The axiom says that if $F$ is a sheaf, and $a,bin F(U)$, and ${U_i}$ is a cover of $U$, then if $a|_{U_i}=b|_{U_i}$ for all $i$, then $a=b$.






          share|cite|improve this answer











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          • $begingroup$
            Why "since $f^sharp(a)$ and $g^sharp(a)$ are equal on an open cover of $U$, we must have that $f^sharp(a)=g^sharp(a)$ since $calO_X$ is a sheaf" holds automatic? I couldn't say this impliance from the definition of the sheaf.
            $endgroup$
            – user621469
            Dec 1 '18 at 22:45
















          2












          $begingroup$

          First, I should hope its clear that if two maps of topological spaces agree on an open cover then they are equal (since maps of topological spaces are functions, so if they agree on any cover of their domain, they are equal as functions).



          Thus the key part is to show that $f^sharp=g^sharp$ as morphisms from $newcommandcalO{mathcal{O}}calO_Yto f_*calO_X=g_*calO_X$.



          Let $V$ be an open subset of $Y$. Let $aincalO_Y(V)$. Let $newcommandinv{^{-1}}U=finv(V)=ginv(V)$. Then $f^sharp(a),g^sharp(a)in calO_X(U)$ by definition. Then ${U_lambdacap U}_{lambdainLambda}$ gives an open cover of $U$, and since $f|_{U_lambda}=g|_{U_lambda}$, we have that $f^sharp(a)|_{Ucap U_lambda}=g^sharp(a)|_{Ucap U_lambda}$. Then since $f^sharp(a)$ and $g^sharp(a)$ are equal on an open cover of $U$, we must have that $f^sharp(a)=g^sharp(a)$ since $calO_X$ is a sheaf. Thus since $a$ and $V$ were arbitrary, $f^sharp=g^sharp$.



          Edit in response to comment:



          To clarify why $f^sharp(a)$ agreeing with $g^sharp(a)$ on an open cover of $U$ implies that they are equal, this is one of the axioms of sheaves. On wiki, this is the locality axiom. The axiom says that if $F$ is a sheaf, and $a,bin F(U)$, and ${U_i}$ is a cover of $U$, then if $a|_{U_i}=b|_{U_i}$ for all $i$, then $a=b$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Why "since $f^sharp(a)$ and $g^sharp(a)$ are equal on an open cover of $U$, we must have that $f^sharp(a)=g^sharp(a)$ since $calO_X$ is a sheaf" holds automatic? I couldn't say this impliance from the definition of the sheaf.
            $endgroup$
            – user621469
            Dec 1 '18 at 22:45














          2












          2








          2





          $begingroup$

          First, I should hope its clear that if two maps of topological spaces agree on an open cover then they are equal (since maps of topological spaces are functions, so if they agree on any cover of their domain, they are equal as functions).



          Thus the key part is to show that $f^sharp=g^sharp$ as morphisms from $newcommandcalO{mathcal{O}}calO_Yto f_*calO_X=g_*calO_X$.



          Let $V$ be an open subset of $Y$. Let $aincalO_Y(V)$. Let $newcommandinv{^{-1}}U=finv(V)=ginv(V)$. Then $f^sharp(a),g^sharp(a)in calO_X(U)$ by definition. Then ${U_lambdacap U}_{lambdainLambda}$ gives an open cover of $U$, and since $f|_{U_lambda}=g|_{U_lambda}$, we have that $f^sharp(a)|_{Ucap U_lambda}=g^sharp(a)|_{Ucap U_lambda}$. Then since $f^sharp(a)$ and $g^sharp(a)$ are equal on an open cover of $U$, we must have that $f^sharp(a)=g^sharp(a)$ since $calO_X$ is a sheaf. Thus since $a$ and $V$ were arbitrary, $f^sharp=g^sharp$.



          Edit in response to comment:



          To clarify why $f^sharp(a)$ agreeing with $g^sharp(a)$ on an open cover of $U$ implies that they are equal, this is one of the axioms of sheaves. On wiki, this is the locality axiom. The axiom says that if $F$ is a sheaf, and $a,bin F(U)$, and ${U_i}$ is a cover of $U$, then if $a|_{U_i}=b|_{U_i}$ for all $i$, then $a=b$.






          share|cite|improve this answer











          $endgroup$



          First, I should hope its clear that if two maps of topological spaces agree on an open cover then they are equal (since maps of topological spaces are functions, so if they agree on any cover of their domain, they are equal as functions).



          Thus the key part is to show that $f^sharp=g^sharp$ as morphisms from $newcommandcalO{mathcal{O}}calO_Yto f_*calO_X=g_*calO_X$.



          Let $V$ be an open subset of $Y$. Let $aincalO_Y(V)$. Let $newcommandinv{^{-1}}U=finv(V)=ginv(V)$. Then $f^sharp(a),g^sharp(a)in calO_X(U)$ by definition. Then ${U_lambdacap U}_{lambdainLambda}$ gives an open cover of $U$, and since $f|_{U_lambda}=g|_{U_lambda}$, we have that $f^sharp(a)|_{Ucap U_lambda}=g^sharp(a)|_{Ucap U_lambda}$. Then since $f^sharp(a)$ and $g^sharp(a)$ are equal on an open cover of $U$, we must have that $f^sharp(a)=g^sharp(a)$ since $calO_X$ is a sheaf. Thus since $a$ and $V$ were arbitrary, $f^sharp=g^sharp$.



          Edit in response to comment:



          To clarify why $f^sharp(a)$ agreeing with $g^sharp(a)$ on an open cover of $U$ implies that they are equal, this is one of the axioms of sheaves. On wiki, this is the locality axiom. The axiom says that if $F$ is a sheaf, and $a,bin F(U)$, and ${U_i}$ is a cover of $U$, then if $a|_{U_i}=b|_{U_i}$ for all $i$, then $a=b$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 1 '18 at 22:59

























          answered Nov 30 '18 at 18:13









          jgonjgon

          14.5k22042




          14.5k22042












          • $begingroup$
            Why "since $f^sharp(a)$ and $g^sharp(a)$ are equal on an open cover of $U$, we must have that $f^sharp(a)=g^sharp(a)$ since $calO_X$ is a sheaf" holds automatic? I couldn't say this impliance from the definition of the sheaf.
            $endgroup$
            – user621469
            Dec 1 '18 at 22:45


















          • $begingroup$
            Why "since $f^sharp(a)$ and $g^sharp(a)$ are equal on an open cover of $U$, we must have that $f^sharp(a)=g^sharp(a)$ since $calO_X$ is a sheaf" holds automatic? I couldn't say this impliance from the definition of the sheaf.
            $endgroup$
            – user621469
            Dec 1 '18 at 22:45
















          $begingroup$
          Why "since $f^sharp(a)$ and $g^sharp(a)$ are equal on an open cover of $U$, we must have that $f^sharp(a)=g^sharp(a)$ since $calO_X$ is a sheaf" holds automatic? I couldn't say this impliance from the definition of the sheaf.
          $endgroup$
          – user621469
          Dec 1 '18 at 22:45




          $begingroup$
          Why "since $f^sharp(a)$ and $g^sharp(a)$ are equal on an open cover of $U$, we must have that $f^sharp(a)=g^sharp(a)$ since $calO_X$ is a sheaf" holds automatic? I couldn't say this impliance from the definition of the sheaf.
          $endgroup$
          – user621469
          Dec 1 '18 at 22:45


















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