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 Classifying Spaces and Classifying Topoi.

A linear order 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 $$ y \leq z \quad iff \quad (y,z) \in \mathcal{O}_x $$
This defines a simplicial topological space $N_{\bullet}$ with $$ N_m(L) = \{(y_0, \dots, y_n) |\ y_0 \leq \dots \leq y_n\ in\ L \} $$ 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.

An augmentation of $L$ is a map of simplicial spaces $N_{\bullet}(L) \to Nerve_{\bullet}(\mathcal{C})$. 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 concordant 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.

The last theorem in Moerdijk's book says:

If $\mathcal{C}$ is a locally contractible topological category and $X$ is a CW-complex, then there is a bijection $$ [X, B\mathcal{C}] \cong Lin_c(X,\mathcal{C}) .$$

There is another result about maps of this kind, which is quite similar and can be found in a paper by Baas, Bökstedt and Kro and says

Assume that $X$ is a CW-complex and $Z_{\bullet}$ is a good simplicial space, then geometric realization induces a bijection $$Con_{Z_{\bullet}}(X) \cong [X, |Z_{\bullet}|] $$

where the elements of $Con_{Z_{\bullet}}(X)$ are given by maps of simplicial spaces of the form $\mathcal{U}_{\bullet} \to Z_{\bullet}$, where $\mathcal{U}_{\bullet}$ denotes the ordered Cech-complex as a simplicial space build from an ordered open cover of $X$. Taking $Z_{\bullet}$ 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 $\mathcal{C}$.

After this lengthy explanation here is my question:

Since you can cook up a linear order from an ordered covering of $X$ there is a map $$Con_{Nerve(\mathcal{C})_{\bullet}}(X) \to Lin_c(X, \mathcal{C})$$ 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?

EDIT: Can I prove directly that the above map is a bijection, i.e. does every augmented linear order come from one induced by an open covering?


1 Answer 1


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 $X \times [0,1]$.

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.

  • $\begingroup$ Thank you! This seems to solve the case, where $\mathcal{C}$ is the category with one object and one morphism (i.e. where there is no augmentation). Or is this also the answer to the general case? I clarified the question a little. $\endgroup$ Mar 8, 2012 at 20:22
  • $\begingroup$ No, it doesn't. I had misunderstood your question. $\endgroup$ Mar 8, 2012 at 21:06

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