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I have a foolish question. But I don't understad it. Could you help me?

Why the following 2 are equivalent? 1. (Y,f) is a universal categorical quotient of X by G. 2. for all affine schemes Y',and morphisms Y' -> Y, if f':X' -> Y' is the base extension,then (Y',f') is a categorical quotient of X' by G.

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As someone who enjoys category theory, but doesn't work in algebraic geometry all that much, it would help if you define your terms. –  David White Oct 27 '11 at 2:44
@David White: Since you enjoy category theory, replace the category of schemes by an arbitrary category and the category of affine schemes by an arbitrary dense subcategory and see what happens. –  Martin Brandenburg Oct 27 '11 at 13:04

2 Answers 2

I assume that by "universal categorical quotient" you mean a morphism $f\colon X\to Y$ that is a categorical quotient, and stays so after any base change $Y'\to Y$.

Then the implication (1)$\Rightarrow$(2) is obvious, and for the other, take any $G$-invariant morphism $g\colon X'\to Z$, and an affine covering $U_i$ of $Y$. Call $f'\colon X'\to Y'$ the base change of $f$.

For any $i$ the map $f'^{-1}(U_i)\to U_i$ will be a categorical quotient by (2) since it is an affine base-change of $f$, and so the restriction $g|_{f'^{-1}(U_i)} \colon f'^{-1} (U_i) \to Z$, being $G$-invariant, will factor through $U_i$. Since this factorization is canonical, those with different $i$'s will be compatible in the intersections (cover it with affines), and in the end you get a (unique) factorization $X'\to Y'\to Z$ for $g$, showing that $f'$ is a categorical quotient.

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The morphism $f:X\to Y$ is a categorical quotient iff for all schemes $Z$, all $G$-invariant morphisms $g:X\to Z$ factor uniquely through a morphism $h:Y\to Z$ such that $g=h$, and $f$ is a universal categorical quotient if for all $Y'\to Y$, the morphism $f':X'=X\times_Y Y'\to Y'$ is a categorical quotient.

The answer to your question is an exercise in glueing. It is obvious that 1 implies 2. Conversely, if $Y'\to Y$ is a morphism then let

  • $\{Y_i\}$ be a covering of $Y$ by open affines,
  • $Y'_i$ and $X'_i$ the preimages in $Y'$ and $X'$,
  • $\{Y'_{ijk}\}$ a covering of $Y'_{ij}:=Y'_i\cap Y'_j$ by open affines,
  • $X'_{ijk}$ the preimage of $Y'_{ijk}$ in $X'$.

By the assumption, the morphisms $X'_i\to Y'_i$ and $X'_{ijk}\to Y'_{ijk}$ are categorical quotients. Therefore if $X'\to Z'$ is a $G$-invariant morphism then for each $i$ there is an induced morphism $h_i:Y'_i\to Z'$ obtained from the invariant map $X'_i\to X'\to Z'$. Moreover the restrictions of $h_i$ and $h_j$ to each open affine $Y'_{ijk}$ coincide, being obtained from the invariant map $X'_{ijk}\to X'\to Z'$, and hence coincide on $Y'_{ij}$. It follows that the $h_i$ glue together into a morphism $Y'\to Z'$ that answers the question.

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Ooops, sorry Mattia, I'm 8 minutes late. But there's no way I can leave a comment on your post, how come ? –  Matthieu Romagny Oct 27 '11 at 9:15
Seems you need at least 50 reputation to leave comments.. don't worry anyway! ;) –  Mattia Talpo Oct 27 '11 at 9:34
I think you accidentally have two accounts, both unregistered. mathoverflow.net/users/5919/matthieu-romagny mathoverflow.net/users/17988/matthieu-romagny You should register an account and post about it here: tea.mathoverflow.net/discussion/605/3/merge-two-user-ids –  Dan Petersen Oct 27 '11 at 9:44
Thanks, Dan. Thus I registered and went to the discussion page you indicate, but then I can't sign in : I'm begin told "Failed to find an account registered with the specified username". –  Matthieu Romagny Oct 27 '11 at 12:59
Now it's all right, thanks again. –  Matthieu Romagny Oct 28 '11 at 8:22

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