this exercise goes as follow :
Let $X$ be an open subscheme of an affine scheme $Y$. Show that the canonical morphism $X \rightarrow Y$ corresponds by the map $\rho$ of proposition 3.25 to the restriction $\mathcal{O}_Y(Y) \rightarrow \mathcal{O}_X(X)$
The map $\rho$ of proposition 3.25 is simply the canonical map sending a morphism of schemes $X \rightarrow Y$ to the induced morphism $\mathcal{O}_Y(Y) \rightarrow \mathcal{O}_X(X)$. Proposition 3.25 states that this map is a bijection is $Y$ is affine.
The thing is, it seems to me that it is just true by definition. What is there to prove here ?
The book never really defines the inclusion morphism of an open subset of a ringed space, but it was always my understanding that if $X$ is an open subset of a ringed space $Y$, the map $\mathcal{O}_Y(U) \rightarrow \mathcal{O}_X(U \cap X) = \mathcal{O}_Y(U \cap X)$ induced by the inclusion morphism is the restriction map in $\mathcal{O}_Y$.
Is there something I'm missing here?
Best Answer
For an open subscheme $X\subset Y$, the canonical morphism is $(i,i^{\sharp}):X\rightarrow Y$, where $i$ is the inclusion of the underlying topological spaces and $i^{\sharp}: \mathcal{O}_Y\rightarrow i_* \mathcal{O}_X$ is given on open sets by the restriction $i^{\sharp}(U):\mathcal{O}_Y (U)\rightarrow i_* \mathcal{O}_X(U)=\mathcal{O}_X(i^{-1}(U))=\mathcal{O}_Y(X\cap U)$. Applying $\rho$ to this map gives us the ring map $i^{\sharp}(Y)$, which is precisely the restriction $\mathcal{O}_Y(Y)\rightarrow \mathcal{O}_Y(X)=\mathcal{O}_X(X)$