My question concerns morphisms between algebraic spaces. I like the definitions of Artin, but I do not see a simple proof of the fact that the composition of two morphisms is a morphism. Could someone explain to me how to get it?

I recall the definitions to see what I exactly mean.

In the book of Artin, an algebraic space is given by an affine scheme $A$, together with a closed equivalence relation $R\subset A\times A$ such that the projections $R\to A$ are étale morphisms.

Then, a morphism from a scheme $X$ to the algebraic space obtained is defined as a "graph" which is a closed subset $W\subset X\times A$ such that the projection $W\to X$ is surjective and étale and such that the pull-back of any point corresponds to an equivalence class of $A$. This naturally induces a map from the points of $X$ to the points of $A$.

A morphism from an algebraic space $A/R$ to another one $A'/R'$ is just a morphism $A\to A'/R'$ which is compatible with the equivalence relation $R$.

Given two morphisms $A/R\to A'/R'$ and $A'/R'\to A''/R''$, the composition should be a morphism. I see how to construct the graph in $A\times A''$, taking the projection of the closed of $A\times A'\times A''$ where the first two elements and the last two elements give elements of the corresponding two graphs of the previous morphisms. However, why is the graph obtained closed and why is the projection étale?

  • $\begingroup$ Why do you need to construct an equivalence relation in the mentioned fibre products ? $\endgroup$ – Damian Rössler Oct 15 '12 at 10:46
  • $\begingroup$ Because by definition a morphism is given by a graph. But I do not see how to compose two... $\endgroup$ – Jérémy Blanc Nov 8 '12 at 18:09
  • $\begingroup$ isn't it much easier to view algebraic spaces as functors? that way a morphism is just a natural transformation. $\endgroup$ – Jacob Bell Nov 29 '12 at 19:08
  • $\begingroup$ It probably works, but I wanted to see it from the definition of Artin. $\endgroup$ – Jérémy Blanc Nov 29 '12 at 23:05

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