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!



Segal maps for Segal precategories
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 .


