# Does Hartshorne *really* not define things like the composition or restriction of morphisms of schemes?

So far as I can tell, Hartshorne's Algebraic Geometry doesn't define the composition of morphisms of schemes, or the restriction of a morphism to an open subset. Of course it's easy enough to define these in several ways, but not having a fixed definition kind of complicates writing up rigorous solutions to some of the more foundational exercises. Am I just missing something obvious here?

-
There is surely only one sensible way of defining composition. What else could there be? –  Zhen Lin Dec 18 '13 at 9:24
I think he's responding to my statement that "it's easy enough to define these in several ways." What I should have said is "it's easy enough to see what the definition of composition should be, and you could define restriction in several equivalent but formally different ways." –  Daniel McLaury Dec 18 '13 at 10:02
(There are comments missing above, making my previous comment sound a bit weird.) –  Daniel McLaury Dec 18 '13 at 10:37
@Hurkyl: I'm sure I'm just being stupid in the wee hours of the morning here, but can you expand on that? What are the two algebras you're talking about? –  Daniel McLaury Dec 18 '13 at 12:27

I agree. On p. 72 locally ringed spaces and their morphisms are defined. Then on p. 73 we have the definition of an isomorphism of locally ringed spaces as a morphism with a two-sided inverse. In principle we need a composition of morphisms for this definition, but Hartshorne doesn't define it there. Instead, he characterizes isomorphisms $(f,f^\#)$ by the property that $f$ is a homeomorphism and $f^\#$ is an isomorphism of sheaves. By the way, with sheaves we have the same problem: On p. 63 isomorphisms of sheaves are defined to be morphisms of sheaves with a two-sided inverse, but not composition of morphisms of sheaves is defined! Finally, on p. 74 morphisms of schemes are defined as morphisms of the underlying locally ringed spaces.

I was a tutor for a lecture on algebraic geometry which was based on the book by Hartshorne. In one exercise one needed to know the precise definition of the composition of two morphisms. Nobody knew it, except for one student. It wasn't explained in the lecture, and the professor didn't even notice that the definition was missing.

One more reason why this book is not the best introduction to algebraic geometry. There are far better introductions, but they often also don't spell out the definition (except of course for EGA I, see Daniel's answer). In Görtz, Wedhorn, Algebraic geometry, the definition is sketched in a remark after Definition 2.29. In Qing Liu, Algebraic Geometry and Arithmetic Curves the defintition is sketched in a remark after Definition 2.20. I could not find the definition in Bosch, Algebraic geometry and commutative algebra; Eisenbud, Harris, The geometry of schemes; Ueno, Algebraic Geometry I.. In most texts such as Vakil, Foundations of algebraic geometry it is just said "there is an obvious notion of composition of morphisms".

-
I'll hold off on accepting this for a little bit in case someone spots a hidden definition in one of the appendices or something weird like that. –  Daniel McLaury Dec 18 '13 at 10:04
So let us just write it down: If $F = (f,f^\#) : (X,\mathcal{O}_X) \to (Y,\mathcal{O}_Y)$ and $G = (g,g^\#) : (Y,\mathcal{O}_Y) \to (Z,\mathcal{O}_Z)$ are morphisms of ringed spaces, then their composition $G \circ F : (X,\mathcal{O}_X) \to (Z,\mathcal{O}_Z)$ is defined by $G \circ F = (g \circ f, g_* (f^\#) \circ g^\#)$. This makes sense since $g_* (f^\#) \circ g^\#$ maps from $\mathcal{O}_Z$ to $g_* (f_* \mathcal{O}_X) = (g \circ f)_* \mathcal{O}_X$. –  Martin Brandenburg Dec 18 '13 at 12:04