maybe this is an idiot question, however I could not solve this after thinking for a while. I added the tag about higher categories simply because of the nature of the question, however this is just a combinatorial exercise.

Let $N$ denote the nerve functor $$N : Cat \rightarrow PrSh (\Delta)$$ ($N(C )_n =Cat([n], C)$) and $$|-|: PrSh(\Delta) \rightarrow Top$$ the geometric realization $|X| = \int^{[n] \in \Delta} X_n \otimes |\Delta^n|$ (the left Kan extension of the obvious realization $\Delta \rightarrow Top$ along the Yoneda embedding).

In page 21 of Lurie's HTT book (see here http://www.math.harvard.edu/~lurie/papers/croppedtopoi.pdf, for instance), he claims that if $J$ is a linearly ordered set and $$P_{i, j} = \{I \subset J; (i,j \in I) \wedge (\forall k \in I : i\leq k \leq j) \}$$ is viewed as a category, then $|N(P_{i, j})| $ is homeomorphic to the cube $[0, 1]^{n}$ where $n$ is the number of elements strictly between $i$ and $j$ (without counting $i$ and $j$).

How can I prove that $|N (P_{i, j})| \cong [0, 1]^n$?

For small values of $n$ ($\leq 3$) this can be easily seen by drawing the nerve. However for general $n$ I'm not sure how to proceed. Maybe counting the numbers of (non-degenerated) $k$-simplices for each $k$ is enough (however I'm not sure about this, is the n-dimensional cube uniquely determined among the simplicial objects by its number of simplices?)

Another way of restating the same question is: consider the simplicial object such that the 1-skeleton is given by the directed graph that is composed by a point that ramifies into $n$ vertices (by adding $n$ edges) and then inductively joins each pair of vertex to a new vertex by adding $2$ edges until there is jut one vertex on the top. Then by gluing the higher simplices along each possible boundary (equivalently, making a 1-coskeletal object), we get a simplicial object. Is this simplicial object an $n$-cube?

Thanks in advance.


1 Answer 1


The following facts might help you:

  1. The category $P_{i,j}$ is isomorphic to $\left[1\right]^{\left]i,j\right[}$, where $\left]i,j\right[ := \{i+1,\dots,j-1\}$, and where $\left[1\right] \in \Delta$.
  2. The nerve functor $N : \mathrm{Cat} \to \mathrm{Set}_{\Delta}$ preserves limits ( - in particular: products! - ), as it is a right adjoint.
  3. The geometric realization functor $|\cdot| : \mathrm{Set}_{\Delta} \to \mathcal{CG}$ commutes with finite products ( - see p. 19 of Lurie's Higher Topos Theory).
  • $\begingroup$ Thanks for your answer. However my problem is exactly your statement 1. Maybe this is a rather trivial, but I'm not seeing the isomorphism. $\endgroup$
    – user40276
    Oct 20, 2015 at 14:22
  • $\begingroup$ The isomorphism sends $I \in P_{i,j}$ to the "functor" $\chi_I : \left]i,j\right[ \to \left[1\right]$ with $\chi_I(k) = 1$ if $k \in I$ and $\chi_I(k) = 0$ if $k \notin I$. $\endgroup$ Oct 20, 2015 at 14:28
  • $\begingroup$ Oh! now I see. How could I not realize this before!? The isomorphism is given by the following rule. Say $J = \{0, 1, …, n+ 1 \}$, then $(a_1, …, a_n) \in [1]^{]i, j[}$ goes to $\{i \in J; i =0 \vee i = n+1 \vee (a_i = 1) \}$ $\endgroup$
    – user40276
    Oct 20, 2015 at 14:29
  • $\begingroup$ ($\left]i,j\right[$ shall be regarded as a discrete category.) $\endgroup$ Oct 20, 2015 at 14:29
  • $\begingroup$ Yes, that's (almost) correct. $\endgroup$ Oct 20, 2015 at 14:32

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