Lemma 26.3.4. Let $f : X \to Y$ be an open immersion of locally ringed spaces. Let $j : V = f(X) \to Y$ be the open subspace of $Y$ associated to the image of $f$. There is a unique isomorphism $f' : X \cong V$ of locally ringed spaces such that $f = j \circ f'$.

Proof. Let $f'$ be the homeomorphism between $X$ and $V$ induced by $f$. Then $f = j \circ f'$ as maps of topological spaces. Since there is an isomorphism of sheaves $f^\sharp : f^{-1}(\mathcal{O}_ Y) \to \mathcal{O}_ X$, there is an isomorphism of rings $f^\sharp : \Gamma (U, f^{-1}(\mathcal{O}_ Y)) \to \Gamma (U, \mathcal{O}_ X)$ for each open subset $U \subset X$. Since $\mathcal{O}_ V = j^{-1}\mathcal{O}_ Y$ and $f^{-1} = f'^{-1} j^{-1}$ (Sheaves, Lemma 6.21.6) we see that $f^{-1}\mathcal{O}_ Y = f'^{-1}\mathcal{O}_ V$, hence $\Gamma (U, f'^{-1}(\mathcal{O}_ V)) \to \Gamma (U, f^{-1}(\mathcal{O}_ Y))$ is an isomorphism for every $U \subset X$ open. By composing these we get an isomorphism of rings

$\Gamma (U, f'^{-1}(\mathcal{O}_ V)) \to \Gamma (U, \mathcal{O}_ X)$

for each open subset $U \subset X$, and therefore an isomorphism of sheaves $f^{-1}(\mathcal{O}_ V) \to \mathcal{O}_ X$. In other words, we have an isomorphism $f'^{\sharp } : f'^{-1}(\mathcal{O}_ V) \to \mathcal{O}_ X$ and therefore an isomorphism of locally ringed spaces $(f', f'^{\sharp }) : (X, \mathcal{O}_ X) \to (V, \mathcal{O}_ V)$ (use Lemma 26.2.2). Note that $f = j \circ f'$ as morphisms of locally ringed spaces by construction.

Suppose we have another morphism $f'' : (X, \mathcal{O}_ X) \to (V, \mathcal{O}_ V)$ such that $f = j \circ f''$. At any point $x \in X$, we have $j(f'(x)) = j(f''(x))$ from which it follows that $f'(x) = f''(x)$ since $j$ is the inclusion map; therefore $f'$ and $f''$ are the same as morphisms of topological spaces. On structure sheaves, for each open subset $U \subset X$ we have a commutative diagram

$\xymatrix@R=5em{ \Gamma (U, f^{-1}(\mathcal{O}_ Y)) \ar[d]_\cong \ar[r]^\cong & \Gamma (U, \mathcal{O}_ X) \\ \Gamma (U, f'^{-1}(\mathcal{O}_ V)) \ar@/^/[ru]^{f'^\sharp } \ar@/_/[ru]_{f''^\sharp } & }$

from which we see that $f'^\sharp$ and $f''^\sharp$ define the same morphism of sheaves. $\square$

Comment #2814 by Samir Canning on

I think there's a missing "is an isomorphism" following

$f^{-1}\mathcal{O}_Y = f'^{-1}\mathcal{O}_V$, hence $\Gamma(U, f'^{-1}(\mathcal{O}_V)) \to \Gamma(U, f^{-1}(\mathcal{O}_Y))$

There are also:

• 2 comment(s) on Section 26.3: Open immersions of locally ringed spaces

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