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definition of isomorphism of ringed spaces



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)definition of morphism of ringed spacesMorphisms between locally ringed spaces and affine schemesWhen is a morphism of $k$-ringed spaces the morphism induced by pullbacks?morphism of ringed spaces glueDefinition of a morphism of locally ringed spacesGeometric intuition behind locality of morphisms of locally ringed spacesOpen vs Closed Immersions of Locally Ringed SpacesLocally ringed spaces and varietiesHow to understand $s_x_sin S$ generates $mathcal F_x$ for every $xin X$?Morphisms of Locally ringed spaces over manifolds.










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$begingroup$


I'm reading Qing Liu's "Algebraic Geometry and Arithmetic Curves." p. 38. I'm not sure I clearly understood the definition of an isomorphism of ringed spaces. In my book, ringed space always mean locally ringed space. The following is the definition of isomorphism from the book:




An isomorphism is an invertible morphism.




So suppose we have two ringed spaces $(X,mathcal O_X)$ and $(Y,mathcal O_Y)$ and a morphism $(f,f^#):(X,mathcal O_X)to(Y,mathcal O_Y)$. To say that $(f,f^#)$ is invertible, does it mean that there is a morphism $(g,g^#):(Y,mathcal O_Y)to(X,mathcal O_X)$ such that both compositions $(f,f^#)circ(g,g^#)$ and $(g,g^#)circ(f,f^#)$ are identity morphisms?










share|cite|improve this question









$endgroup$











  • $begingroup$
    Yes. This definition of isomorphism is completely general.
    $endgroup$
    – KReiser
    Apr 2 at 18:26















0












$begingroup$


I'm reading Qing Liu's "Algebraic Geometry and Arithmetic Curves." p. 38. I'm not sure I clearly understood the definition of an isomorphism of ringed spaces. In my book, ringed space always mean locally ringed space. The following is the definition of isomorphism from the book:




An isomorphism is an invertible morphism.




So suppose we have two ringed spaces $(X,mathcal O_X)$ and $(Y,mathcal O_Y)$ and a morphism $(f,f^#):(X,mathcal O_X)to(Y,mathcal O_Y)$. To say that $(f,f^#)$ is invertible, does it mean that there is a morphism $(g,g^#):(Y,mathcal O_Y)to(X,mathcal O_X)$ such that both compositions $(f,f^#)circ(g,g^#)$ and $(g,g^#)circ(f,f^#)$ are identity morphisms?










share|cite|improve this question









$endgroup$











  • $begingroup$
    Yes. This definition of isomorphism is completely general.
    $endgroup$
    – KReiser
    Apr 2 at 18:26













0












0








0





$begingroup$


I'm reading Qing Liu's "Algebraic Geometry and Arithmetic Curves." p. 38. I'm not sure I clearly understood the definition of an isomorphism of ringed spaces. In my book, ringed space always mean locally ringed space. The following is the definition of isomorphism from the book:




An isomorphism is an invertible morphism.




So suppose we have two ringed spaces $(X,mathcal O_X)$ and $(Y,mathcal O_Y)$ and a morphism $(f,f^#):(X,mathcal O_X)to(Y,mathcal O_Y)$. To say that $(f,f^#)$ is invertible, does it mean that there is a morphism $(g,g^#):(Y,mathcal O_Y)to(X,mathcal O_X)$ such that both compositions $(f,f^#)circ(g,g^#)$ and $(g,g^#)circ(f,f^#)$ are identity morphisms?










share|cite|improve this question









$endgroup$




I'm reading Qing Liu's "Algebraic Geometry and Arithmetic Curves." p. 38. I'm not sure I clearly understood the definition of an isomorphism of ringed spaces. In my book, ringed space always mean locally ringed space. The following is the definition of isomorphism from the book:




An isomorphism is an invertible morphism.




So suppose we have two ringed spaces $(X,mathcal O_X)$ and $(Y,mathcal O_Y)$ and a morphism $(f,f^#):(X,mathcal O_X)to(Y,mathcal O_Y)$. To say that $(f,f^#)$ is invertible, does it mean that there is a morphism $(g,g^#):(Y,mathcal O_Y)to(X,mathcal O_X)$ such that both compositions $(f,f^#)circ(g,g^#)$ and $(g,g^#)circ(f,f^#)$ are identity morphisms?







algebraic-geometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 2 at 10:30









zxcvzxcv

1919




1919











  • $begingroup$
    Yes. This definition of isomorphism is completely general.
    $endgroup$
    – KReiser
    Apr 2 at 18:26
















  • $begingroup$
    Yes. This definition of isomorphism is completely general.
    $endgroup$
    – KReiser
    Apr 2 at 18:26















$begingroup$
Yes. This definition of isomorphism is completely general.
$endgroup$
– KReiser
Apr 2 at 18:26




$begingroup$
Yes. This definition of isomorphism is completely general.
$endgroup$
– KReiser
Apr 2 at 18:26










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