Semi-simplicial versus simplicial sets (and simplicial categories) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T17:13:34Z http://mathoverflow.net/feeds/question/75094 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75094/semi-simplicial-versus-simplicial-sets-and-simplicial-categories Semi-simplicial versus simplicial sets (and simplicial categories) Peter Arndt 2011-09-10T12:58:45Z 2011-09-18T11:05:42Z <p>Hi,</p> <p>Let's denote by "semi-simplicial set" a simplicial set without degeneracies, i.e. a contravariant functor from the category $\Delta_{inj}$ of finite linearly ordered sets and order preserving injections to sets (this is also known as $\Delta$-set).</p> <p>We have an inclusion functor $j: \Delta_{inj} \rightarrow \Delta$ giving us an adjunction between semi-simplicial sets and simplicial sets: The right adjoint is $j^*: Set^{\Delta^{op}} \rightarrow Set^{\Delta_{inj}^{op}}$ given by precomposition with $j$ (i.e. forgetting degeneracies), the left adjoint is $j_!:Set^{\Delta_{inj}^{op}} \rightarrow Set^{\Delta^{op}}$ given by left Kan extension (i.e. taking realization in simplicial sets). It is known from the article referred to in <a href="http://mathoverflow.net/questions/57653/degeneracies-for-semi-simplicial-kan-complexes/57655#57655" rel="nofollow">this</a> MO-answer that boosting a semi-simplicial set up to a simplicial one and then forgetting degeneracies again results in an equivalent semi-simplicial set (in the sense that their realizations are equivalent).</p> <p>I would like to know about the other direction: Does anyone know conditions on a simplicial set $K$ which ensure that the map $j_!j^*K \rightarrow K$ is a weak equivalence?</p> <p>I am especially interested in the case where $K$ is the nerve of a category.</p> <hr> <p>A related question is the following: Given a Reedy cofibrant semi-cosimplicial object in the category of simplicial categories, which is equivalent to the underlying semi-cosimplicial object of the usual cosimplicial object, one can by the usual yoga set up an adjunction between semi-simplicial sets and simplicial categories. </p> <p>Starting with a simplicial set $K$ one can view it as a semi-simplicial set, then produce a simplicial category. I would like to know under which conditions on $K$ this simplicial category is equivalent to the simplicial category which one gets by applying the usual functor from simplicial sets to simplicial categories (the left adjoint to the coherent nerve) directly to $K$. Again the case of biggest interest for me is when $K$ is the nerve of a category.</p> http://mathoverflow.net/questions/75094/semi-simplicial-versus-simplicial-sets-and-simplicial-categories/75101#75101 Answer by Karol Szumiło for Semi-simplicial versus simplicial sets (and simplicial categories) Karol Szumiło 2011-09-10T13:58:55Z 2011-09-10T13:58:55Z <p>I haven't thought about your second question, but the counit <code>$\varepsilon_K:j_!j^*K\to K$</code> is always a weak equivalence. It follows from the fact that the unit is a weak equivalence. First, observe that a map of simplicial sets <code>$f:X\to Y$</code> is a weak equivalence if and only if <code>$j^*f$</code> is a weak equivalence, because <code>$|j^*X|$</code> is the "fat geometric realization" of <code>$X$</code>, which is naturally homotopy equivalent to the usual geometric realization. Since the unit is always a weak equivalence, it follows from one of the triangular identities that <code>$j^*\varepsilon_K$</code> is a weak equivalence and by the above observation so is <code>$\varepsilon_K$</code>.</p> http://mathoverflow.net/questions/75094/semi-simplicial-versus-simplicial-sets-and-simplicial-categories/75740#75740 Answer by Jacob Lurie for Semi-simplicial versus simplicial sets (and simplicial categories) Jacob Lurie 2011-09-18T11:05:42Z 2011-09-18T11:05:42Z <p>The map $j_! j^{\ast} K \rightarrow K$ is never a Joyal equivalence unless $K$ is empty. For example, if $K = \Delta^{0}$, then $j_{!} j^{\ast} K$ is the nerve of the category with one object $X$ and a single nonidentity morphism $e: X \rightarrow X$ satisfying $e^2 = e$. You can think of $j_{!} j^{\ast} K$ as "obtained from $K$ by freely adjoining new identity morphisms".</p>