A short writeup of the proof that local morphisms of ringed spaces are given by composition.

1. LOCAL MORPHISMS ARE GIVEN BY COMPOSITION
HEINRICH HARTMANN
Recall the following basic definition.
Definition 0.1. A ringed space (X, O) consists of a topological space X,
and O a sheaf of rings on X.
We call (X, O) a locally ringed space if for all x ∈ X the stalk Ox is a local
ring. We denote the maximal ideal of Ox by mx.
A morphism of ringed spaces f : (X, OX) → (Y, OY ) is pair (f, f†) consisting
of a continues map f : X → Y and f† : OY → f∗OX a morphism of sheaves
of rings on X.
A morphism of locally ringed spaces is called local if for all x ∈ X it is f†
x :
Of(x) → Ox a local homomorphism of local rings (i.e. (f†)−1(mx) = mf(x)).
Definition 0.2. We say that a ringed space (X, O) is function-like if the
following conditions hold.
(1) It is O contained in the sheaf CX of continues functions to C and
the inclusion is a homomorphism of (sheaves of) rings.
(2) The constant functions CX ⊂ CX are contained in O.
(3) It is (X, O) local.
Lemma 0.3. Let (X, O) be function-like. Then the maximal ideal mx ⊂ Ox
at a point x ∈ X is given by mx = mC
x ∩Ox where mC
x ⊂ CX,x is the maximal
ideal of functions vanishing at x.
Proof. Consider the composition φ : Ox ⊂ CX,x
evx
−→ C. It is φ a ring ho-
momorphism with kernel mC
x ∩ Ox. It is surjective since Ox contains the
constant functions. Since C is a field the kernel is a maximal ideal of Ox.
But a local ring has only one maximal ideal.
Proposition 0.4. Let (X, OX) and (Y, OY ) be function-like ringed spaces.
For a morphism of ringed spaces f : (X, OX) → (Y, OY ) are equivalent:
(1) It is f local and f† is C-linear.
(2) The morphism f† is given by composing functions: f∗ : g → g ◦ f.
Proof. Let (2) be true. The C-linearity of f∗ is clear. We have to show that
f∗ is local, i.e. (f∗)−1(mx) = mf(x). This is equivalent to: f∗(g)(x) = 0 if
and only if g(f(x)) = 0 - which is obvious.
Let (1) be true. Let g ∈ Oy and choose a representative g ∈ OY (U)
on an open set y ∈ U ⊂ Y . We have to show f†(g)(x ) = g(f(x )) for
all x ∈ f−1(U). Set c = f†(g)(x ) and regard c as a constant function
on Y resp. X. It is then f†(g) − c ∈ mx since this function vanishes at
x and lies in OX we can apply the above lemma. By C-linearity we have
f†(g) − c = f†(g − c) and by locality it follows g − c ∈ mf(x ) and hence
g(f(x )) = c.
Date: 29.04.2010.
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