Classifying spaces of topological categories - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T13:40:49Z http://mathoverflow.net/feeds/question/90604 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90604/classifying-spaces-of-topological-categories Classifying spaces of topological categories Ulrich Pennig 2012-03-08T17:49:30Z 2012-03-08T20:20:23Z <p>I try to understand and compare the following facts about maps into classifying spaces of topological categories. Consider first the following definition cited from Moerdijk's <em>Classifying Spaces and Classifying Topoi</em>.</p> <p>A <em>linear order</em> over a topological space $X$ is a (set-valued) sheaf $L$ on $X$ (also known as etale topological space) together with a subsheaf $\mathcal{O} \subseteq L\ \times_X\ L$, such that for each point $x \in X$ the stalk $L_x$ is non-empty and linearly ordered by the relation $$y \leq z \quad iff \quad (y,z) \in \mathcal{O}_x$$<br> This defines a simplicial topological space <code>$N_{\bullet}$</code> with <code>$$N_m(L) = \{(y_0, \dots, y_n) |\ y_0 \leq \dots \leq y_n\ in\ L \}$$</code> Now let $\mathcal{C}$ be a category internal to topological spaces. A morphism of linear orders is a map of sheaves which restricts to an order-preserving map on the stalks. </p> <p>An <em>augmentation</em> of $L$ is a map of simplicial spaces <code>$N_{\bullet}(L) \to Nerve_{\bullet}(\mathcal{C})$</code>. A morphism of augmented linear orders is a morphism of linear orders that respects the augmentations. The category of augmented linear orders will be denoted by $Lin(X,\mathcal{C})$. Two elements of this category will be called <em>concordant</em> if there is an augmented linear order over $X \times [0,1]$, such that it restricts to the given ones over the endpoints. Denote by $Lin_c(X,\mathcal{C})$ the concordance classes.</p> <p>The last theorem in Moerdijk's book says:</p> <blockquote> <blockquote> <p>If $\mathcal{C}$ is a locally contractible topological category and $X$ is a CW-complex, then there is a bijection $$[X, B\mathcal{C}] \cong Lin_c(X,\mathcal{C}) .$$</p> </blockquote> </blockquote> <p>There is another result about maps of this kind, which is quite similar and can be found in a <a href="http://arxiv.org/abs/math/0612549" rel="nofollow">paper</a> by Baas, Bökstedt and Kro and says </p> <blockquote> <blockquote> <p>Assume that $X$ is a CW-complex and $Z_{\bullet}$ is a good simplicial space, then geometric realization induces a bijection <code>$$Con_{Z_{\bullet}}(X) \cong [X, |Z_{\bullet}|]$$</code></p> </blockquote> </blockquote> <p>where the elements of $Con_{Z_{\bullet}}(X)$ are given by maps of simplicial spaces of the form <code>$\mathcal{U}_{\bullet} \to Z_{\bullet}$</code>, where $\mathcal{U}_{\bullet}$ denotes the ordered Cech-complex as a simplicial space build from an ordered open cover of $X$. Taking <code>$Z_{\bullet}$</code> to be the nerve of $\mathcal{C}$ we get a bijection $$Lin_c(X, \mathcal{C}) \cong Con_{Nerve(\mathcal{C})_{\bullet}}(X)$$ for nice enough topological categories $\mathcal{C}$.</p> <p>After this lengthy explanation here is my question:</p> <blockquote> <blockquote> <p>Since you can cook up a linear order from an ordered covering of $X$ there is a map <code>$$Con_{Nerve(\mathcal{C})_{\bullet}}(X) \to Lin_c(X, \mathcal{C})$$</code> which should fit into the setup described above. So somehow it seems to be the case that every linear order is concordant to one coming from an ordered covering. Is this true? Can this be proven directly? </p> <p>EDIT: Can I prove directly that the above map is a bijection, i.e. does every <em>augmented</em> linear order come from one induced by an open covering? </p> </blockquote> </blockquote> http://mathoverflow.net/questions/90604/classifying-spaces-of-topological-categories/90615#90615 Answer by Oscar Randal-Williams for Classifying spaces of topological categories Oscar Randal-Williams 2012-03-08T19:23:48Z 2012-03-08T19:23:48Z <p>Let $L_0$ and $L_1$ be two linear orders over $X$ (I think of these as being the etale space). Let $L'_0 = L_0 \times [0,1)$, which is an etale space over $X \times [0,1]$; let $L'_1 = L_1 \times (0,1]$ which is another etale spacer over $X \times [0,1]$.</p> <p>Define $L = L'_0 \coprod L'_1$, with the order relation that everything in $L'_1$ is larger than everything in $L'_0$ (in each fibre). This is a linear order over $X \times [0,1]$, so a concordance from $L_0$ to $L_1$. This shows that any two linear orders are concordant. In particular, the result you wanted follows immediately.</p>