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definition of isomorphism of ringed spaces

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?

algebraic-geometry

share|cite|improve this question

asked Apr 2 at 10:30

zxcvzxcv

1919

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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?

algebraic-geometry

share|cite|improve this question

asked Apr 2 at 10:30

zxcvzxcv

1919

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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?

algebraic-geometry

share|cite|improve this question

asked Apr 2 at 10:30

zxcvzxcv

1919

$endgroup$

share|cite|improve this question

asked Apr 2 at 10:30

zxcvzxcv

1919

share|cite|improve this question

asked Apr 2 at 10:30

zxcvzxcv

1919

asked Apr 2 at 10:30

zxcvzxcv

1919

asked Apr 2 at 10:30

zxcvzxcv

1919