A Segal precategory is just a simplicial space $X:\Delta^{op} \to sSet$ such that its $0$-th space is discrete (i.e. constant). A Segal category is defined everywhere in the literature as a Segal precategory satisfying the Segal condition.

Namely, we want the natural maps $$\phi_k:X_k \to X_1 \times_{X_0} \cdots \times_{X_0} X_1$$ to be weak equivalences $\forall k\geq2$.

There are several reasons for requiring the codomain to be a homotopy limit, and in the analogous case of Segal spaces we see immediately that this holds.

My question is: why does the discreteness of $X_0$ imply the codomain of $\phi_k$ is a homotopy limit? I guess it may be useful to know that $X_0$ is automatically fibrant.

Thanks in advance for any hint!


For any $a,b \in X_0$, let $X_1(a,b)$ denote the subspace of $X_1$ lying over $(a,b) \in X_0 \times X_0$. Since $X_0$ is discrete, this iterated fiber product breaks up as a disjoint union of product spaces: $$ \bigcup_{(a_0, \ldots, a_k)} X_1(a_{k-1}, a_k) \times \cdots \times X_1(a_0,a_1) $$ Generally, in order for fiber products to be homotopy fiber products we usually demand that one of the maps be a fibration. However, finite products of simplicial sets preserve all weak equivalences and so this is automatically a homotopy fiber product.

If you don't like this reasoning, you can move to simplicial topological spaces instead, where every map $Y \to S$ is a fibration when $S$ is discrete .

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    $\begingroup$ I got your first argument, but I think you also have to prove the homotopy invariance of the construction $X_1(a,b)$, which again follows by discreteness of the $0$-th space (so that, in particular, it does not hold in general). $\endgroup$ – Edoardo Lanari May 29 '15 at 19:08
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    $\begingroup$ @EdoardoLanari Yes, absolutely, you need that a decomposition into disconnected components preserves equivalences. $\endgroup$ – Tyler Lawson May 29 '15 at 19:42

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