most recent 30 from http://mathoverflow.net 2013-05-19T23:31:44Z http://mathoverflow.net/feeds/question/65890 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65890/name-for-enrichment-with-hom1-a-full-functor Lamont C 2011-05-24T19:18:27Z 2011-05-24T19:18:27Z

<p>Let C be a V-enriched category and 1 be a terminal object of C. V is not necessarily a closed category, and C does not necessarily have an internal hom (nor is C even necessarily a monoidal category).</p> <blockquote> <p>Is there a term for such an enrichment in which Hom(1,-):C->V is a full functor?</p> </blockquote> <p>and</p> <blockquote> <p>Are there any non-obvious sufficient conditions which imply that Hom(1,-):C->V is full?</p> </blockquote> <p>Background: in a certain sense, when Hom(1,-):C->V is full, it means that from V's perspective the only way to turn a morphism f:1->A into a morphism 1->B is to find some g:A->B and compose $g\circ f$. This is a consequence of the fact that every V-map C(1,A)->C(1,B) arises as Hom(1,g) for some C-map g. So in this scenario, C "knows about" all the ways of turning a 1->A into a 1->B (or, at least knows about all the ways that V knows about).</p> <p>I am interested in learning more about the properties of these sorts of enrichments, but specifically without assuming that V is a closed category (which seems to be the case that gets the most attention in the literature I've been able to find).</p>