universal categorical quotient - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T05:59:06Z http://mathoverflow.net/feeds/question/79224 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79224/universal-categorical-quotient universal categorical quotient google 2011-10-27T02:32:19Z 2012-10-28T12:26:13Z <p>I have a foolish question. But I don't understad it. Could you help me?</p> <p>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.</p> http://mathoverflow.net/questions/79224/universal-categorical-quotient/79247#79247 Answer by Mattia Talpo for universal categorical quotient Mattia Talpo 2011-10-27T09:05:42Z 2011-10-27T09:05:42Z <p>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 $Y'\to Y$.</p> <p>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 $f$.</p> <p>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.</p> http://mathoverflow.net/questions/79224/universal-categorical-quotient/79248#79248 Answer by Matthieu Romagny for universal categorical quotient Matthieu Romagny 2011-10-27T09:14:09Z 2011-10-27T09:14:09Z <p>The morphism <code>$f:X\to Y$</code> is a <em>categorical quotient</em> iff for all schemes <code>$Z$</code>, all <code>$G$</code>-invariant morphisms <code>$g:X\to Z$</code> factor uniquely through a morphism <code>$h:Y\to Z$</code> such that <code>$g=h$</code>, and <code>$f$</code> is a <em>universal categorical quotient</em> if for all <code>$Y'\to Y$</code>, the morphism <code>$f':X'=X\times_Y Y'\to Y'$</code> is a categorical quotient.</p> <p>The answer to your question is an exercise in glueing. It is obvious that 1 implies 2. Conversely, if <code>$Y'\to Y$</code> is a morphism then let</p> <ul> <li><code>$\{Y_i\}$</code> be a covering of <code>$Y$</code> by open affines,</li> <li><code>$Y'_i$</code> and <code>$X'_i$</code> the preimages in <code>$Y'$</code> and <code>$X'$</code>,</li> <li><code>$\{Y'_{ijk}\}$</code> a covering of <code>$Y'_{ij}:=Y'_i\cap Y'_j$</code> by open affines,</li> <li><code>$X'_{ijk}$</code> the preimage of <code>$Y'_{ijk}$</code> in <code>$X'$</code>.</li> </ul> <p>By the assumption, the morphisms <code>$X'_i\to Y'_i$</code> and <code>$X'_{ijk}\to Y'_{ijk}$</code> are categorical quotients. Therefore if <code>$X'\to Z'$</code> is a <code>$G$</code>-invariant morphism then for each <code>$i$</code> there is an induced morphism <code>$h_i:Y'_i\to Z'$</code> obtained from the invariant map <code>$X'_i\to X'\to Z'$</code>. Moreover the restrictions of <code>$h_i$</code> and <code>$h_j$</code> to each open affine <code>$Y'_{ijk}$</code> coincide, being obtained from the invariant map <code>$X'_{ijk}\to X'\to Z'$</code>, and hence coincide on <code>$Y'_{ij}$</code>. It follows that the <code>$h_i$</code> glue together into a morphism <code>$Y'\to Z'$</code> that answers the question.</p>