Exercise 3.10 from Hartshorne

Question

This is Exercise 3.10(a) from Hartshorne. Can someone verify my solution? Thanks.

Exercise 3.10(a): If $f:X\to Y$ is a morphism, and $y\in Y$ is a point, show that $\operatorname{sp}(X_y)$ is homeomorphic to $f^{-1}(y)$ as topological spaces with the induced topology.

Here, $X_y = X\times_{Y} \operatorname{Spec} k(y)$.

Let $f^{-1}(y)\to X$ be the inclusion map, and $f^{-1}(y)\to \operatorname{Spec}k(y) = \{ \text{pt} \}$ be the constant map. These maps commute with their natural maps into $Y$. Therefore, $f^{-1}(y)$ factors through the fibre product by the universal property, and we obtain the map in one direction $$ \gamma: f^{-1}(y)\to X_y $$

However, it is clear that the natural projection $$ X_y\to X $$ has image in $f^{-1}(y)$ by looking at the commutative diagram of the universal property, and hence it defines $\gamma^{-1}$, making $\gamma$ a homeomorphism.

EDIT 1: I suppose I'll have to first treat $f^{-1}(y)$ as some kind of subscheme of $X$ in order to apply the universal property, i.e. I should probably give it a natural structure sheaf (if it is possible).

Answer 1

Your edit is correct: you have not specified a scheme structure on $f^{-1}(y)$, so the maps out of it to $X$ and $\operatorname{Spec} k(y)$ are not maps of schemes. Therefore you cannot use the universal property of the fiber product in the category of schemes because you don't have scheme morphisms, and your effort does not work out.

Instead, I would suggest tackling the affine case first: if $X\to Y$ is $\operatorname{Spec} A\to \operatorname{Spec} R$, what is $f^{-1}(y)$? What is $\operatorname{Spec} A \times_{\operatorname{Spec} R} \operatorname{Spec} k(y)$? Can you relate the two? Full details are available in a solution here, though I'd encourage you to give it a try on your own before reading that.