most recent 30 from http://mathoverflow.net 2013-05-19T18:44:49Z http://mathoverflow.net/feeds/question/38593 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38593/when-does-a-morphism-of-sketches-induce-an-adjunction-between-their-set-models David Spivak 2010-09-13T15:56:07Z 2010-09-13T15:56:07Z

<p>Sketches come in many flavors, according to "what the user may specify." There are finite product sketches where the user may specify a choice of "discrete cones" to be limits. More generally, there are limit sketches where the user may specify a choice of cones of arbitrary shape to be limits. There are similarly coproduct and colimit sketches, as well as combinations. One can allow the user to specify only finite limits and infinite colimits, for example. Most generally (in the context of this mathoverflow question), a sketch may specify any set of small colimits or small limits.</p> <p>Thus there are many restrictions one can put on "what the user can specify." Let's call each such restriction a <em>type of sketch</em>. For each type of sketch ${\bf T}$, we will write ${\bf T}-Sketch$ to denote the category of sketches of type ${\bf T}$ and the sketch maps between them.</p> <p>The category $Set$ of sets sits inside of ${\bf T}-Sketch$ for any type of sketch ${\bf T}$. A $Set$-model of a sketch $S$ is a sketch map $S\to Set$, i.e. a map which preserves all the specified structure. Let $S-Set$ denote the category of $Set$-models of $S$.</p> <p>For any type of sketch ${\bf T}$, and any morphism $f\colon S\to S'$ of ${\bf T}$-sketches, there is a functor $$f^\ast\colon S'-Set\longrightarrow S-Set$$ given by composition.</p> <p>Q1: For what types of sketch does $f^\ast$ have a right adjoint? </p> <p>Q2: For what types of sketch does $f^\ast$ have a left adjoint?</p> <p>It seems that the types of sketch ${\bf T}$ itself forms a poset under inclusion (i.e. "only filtered colimits" is a subtype of "all colimits"). This <em>poset of sketch-types</em> has meets and joins.</p> <p>Q3: Can the answer to Q1 and/or Q2 be phrased in terms of the poset of sketch-types? For example, is the sub-poset of sketch-types for which $f^\ast$ has a right adjoint closed under meets? Joins?</p>