 Solution

ID: past-exam-of-the-introduction-to-schemes-course-of-the-university-of-oxford/2017/1/a/solution

Apply the surjectivity of

f (Y)

1_{Y}

g : Y \to X

satisfying

f \circ g = 1_{Y}

. To obtain the other identity, compare

g \circ f

and

1_{X}

X (X)

. Their images under

f (X)

agree, since

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

The injectivity of

f (X)

therefore gives

g \circ f = 1_{X}

. Thus

f is an isomorphism of schemes, with inverse g .

Only surjectivity for the test scheme

Y

and injectivity for the test scheme

X

were needed. This is the isomorphism-detection argument behind the Yoneda lemma for the functor of points.

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