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 Incoming links: Gluing morphisms of schemes

Yes. First, the hypothesis implies that

f (T)

is injective for every scheme

T

, even when

T

is not an affine scheme. Indeed, suppose

a, b : T \to X

satisfy

f \circ a = f \circ b

T = ⋃_{λ} V_{λ}

. On each

V_{λ}

, injectivity of

f (V_{λ})

gives

a ∣_{V_{λ}} = b ∣_{V_{λ}}

a = b

Y = ⋃_{i} U_{i}

, and denote its inclusion maps by

j_{i} : U_{i} ↪ Y

. Surjectivity of

f (U_{i})

g_{i} : U_{i} \to X

with

f \circ g_{i} = j_{i}

. On

U_{i} \cap U_{j}

, the two restrictions have the same composite with

f

. By the injectivity just established for arbitrary test schemes,

g_{i} ∣_{U_{i} \cap U_{j}} = g_{j} ∣_{U_{i} \cap U_{j}} .

In particular, the overlaps need not be affine schemes. Gluing morphisms of schemes supplies

g : Y \to X

with

g ∣_{U_{i}} = g_{i}

, and

f \circ g = 1_{Y}

because this holds on every

U_{i}

Finally,

f \circ (g \circ f) = f = f \circ 1_{X} .

Injectivity of

f (X)

, already proved without an affineness assumption, gives

g \circ f = 1_{X}

. Therefore

f is an isomorphism of schemes .

This proves the relevant instance of affine test schemes determine scheme morphisms: the restricted functor of points still retains the information needed to construct the inverse.