simplicial spaces without degeneracies - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T07:30:47Z http://mathoverflow.net/feeds/question/91634 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91634/simplicial-spaces-without-degeneracies simplicial spaces without degeneracies Ulrich Pennig 2012-03-19T16:46:41Z 2012-03-20T18:05:59Z <p>Suppose I have a simplicial space $X_{\bullet}$ <em>without</em> degeneracies (sometimes called semi-simplicial space or incomplete simplicial space). There still is a geometric realization $\lVert X \rVert$ of $X_{\bullet}$, which only uses the face maps. What properties does this realization have?</p> <ul> <li>Does it still preserve products, i.e. is $\lVert X \times Y \rVert$ still (weakly?) homotopy equivalent to $\lVert X \rVert \times \lVert Y \rVert$? </li> <li>Do levelwise (weak) equivalences still induce a (weak) homotopy equivalence of the geometric realizations?</li> </ul> <p>I know of the paper "Categories and Cohomology Theories" by Segal, where he defines this "fat" geometric realization in the appendix. Unfortunately, he proves the above properties by a comparison with another construction that takes the degeneracies into account. Is this the only way to prove this? Do the properties fail in case there are no degeneracies?</p> http://mathoverflow.net/questions/91634/simplicial-spaces-without-degeneracies/91640#91640 Answer by Hiro Lee Tanaka for simplicial spaces without degeneracies Hiro Lee Tanaka 2012-03-19T17:08:17Z 2012-03-20T18:05:59Z <p>[ I now realize that the OP is asking about functors in the image of the forgetful functor from simplicial spaces to semisimplicial spaces, though I focus my answer mainly on a general semisimplicial space. I'll keep the answer up anyway in case it's helpful.</p> <p>To clarify, by a simplicial blah, I'm as usual talking about a functor $\Delta^{op} \to Blahs$. A semisimplicial blah is given by a functor $\Delta_0^{op} \to Blahs$, where $\Delta_0$ is the non-full subcategory where we discard the non-identity surjections.]</p> <p>The degeneracies are critical when considering products--it is in fact false that the geometric realization commutes with products (up to homotopy equivalence) without the degenerate simplices. A simple example is given by taking the product of $S^1$ with itself---if you take the one-vertex, one-edge semisimplicial structure, you'll never recover the torus via geometric realization. (An obvious problem arises when now your higher simplices $X_n$ can equal the empty set.) Note this is an example in semisimplicial sets, not even spaces. In short, the functor $||\bullet||: semiSSpace \to Spaces$ is not well-behaved with respect to products.</p> <p>However, these "incomplete" simpicial sets/spaces arise really naturally. For instance, a lot of categories may not have a natural unit/identity, so there are no degeneracy maps in the nerve of the category---these are modeled most easily by "semi"simplicial sets/spaces. But so long as you're not taking products, the theory of such things (I think) work out just fine. You can still talk about the classifying space of a non-unital category by taking geometric realization of the corresponding semisimplicial set (i.e., its nerve).</p> <p>As for your second question, I'm fairly certain that what you say is correct--if you have two semisimplicial spaces $Y_\bullet, X_\bullet$ with a natural transformation that induces level-wise equivalences, then the geometric realizations will be (weakly) homotopy equivalent. I think you can see this by taking a filtration by simplicial index and seeing that the associated gradeds are equivalent.</p> <p>Given a semisimplicial space, there are some ways to get an actual simplicial space (this is akin to formally adding units in a category, or to an algebra) but I'm not sure how well-behaved this functor is. </p> <p>However, the composite functor $$SSpace \to semiSSpace \to Spaces$$ (forget, and then realize) will satisfy all the properties you asked for, mainly because the fat geometric realization doesn't differ in homotopy type from the usual geometric realization (provided the simplicial space is good, in the sense of Segal!).</p> <p>As always if anybody has questions or more enlightening comments, please share.</p> <p>Hiro</p> http://mathoverflow.net/questions/91634/simplicial-spaces-without-degeneracies/91668#91668 Answer by Oscar Randal-Williams for simplicial spaces without degeneracies Oscar Randal-Williams 2012-03-19T21:01:20Z 2012-03-19T21:01:20Z <p>In brief:</p> <p>For your first question, no. Let $X_\bullet$ be any semi-simplicial space and $Y_\bullet$ have a point in degree zero and be empty in every other degree. Then $\vert X_\bullet \times Y_\bullet \vert = X_0$, which will not usually be equivalent to $\vert X_\bullet \vert$.</p> <p>For your second question, yes. This is always true for semi-simplicial spaces, and is not always true for simplicial spaces (It is in the case where the thick and thin realisations are equivalent, of course). To see this you prove that the maps on $k$-skeleta $\vert X_\bullet \vert^{(k)} \to \vert Y_\bullet \vert^{(k)}$ are equivalences by induction on $k$, using the push-out description of the $k$-skeleton from the $(k-1)$-skeleton, and the fact that $$X_k \times \partial \Delta^k \to X_k \times \Delta^k$$ is a cofibration so that it is a homotopy push-out. Then you use that $\vert X^\bullet \vert = \mathrm{colim} \vert X^\bullet \vert^{(k)}$ and that each $\vert X^\bullet \vert^{(k-1)} \to \vert X^\bullet \vert^{(k)}$ is a cofibration so this is a homotopy colimit.</p>