**Question:**  

Let $X$ be a topological space, $U\_\*\rightarrow X$ be an augmented simplicial space, and let $M\_n(U\_\*)$ be the n-th matching object computed in $sTop$ while $M\_n^X(U\_\*)$ denotes the $n$-th matching object computed in the category $s(Top \downarrow X)$ of simplicial spaces over $X$.


How does one see that for $n \geq 2$, $M\_n^X(U\_\*)$ is isomorphic to $M\_n(U\_\*)$?  

(This is claimed in the paragraph following Def. 4.1 of [DI] but is not obvious to me.)


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**Background definitions:**  

For a category $C$, let $sC$ denote the category of simplicial objects in $C$ and and $s\_{\leq n}C$ denote the category of truncated simplicial objects in $C$ of dimension $n$.  

Let $$sk\_n: sC \rightarrow s\_{\leq n}C$$ be the forgetful functor and let $cosk\_n : s\_{\leq n}C \rightarrow sC$ denote its right adjoint.  

Finally, for $U\_\* \in sC$ , define the $\textbf{n-th matching object, } M\_n(U\_\*),$ to be the $n$-th object of $cosk\_{n-1}(sk\_{n-1}(U\_\*))$. 


**Reference:**  

[DI] Dugger-Isaksen "Hypercovers in topology"  

(<https://arxiv.org/abs/math/0111287>)



Matching objects and hypercovers in topology
This follows from the *construction* of $(\operatorname{cosk}\_m(X\_\bullet))\_n$ as a limit indexed by $(\Delta/[n])^{\operatorname{op}}\_{\leq m}$ (in the non-augmented case) or $(\Delta\_a/[n])^{\operatorname{op}}\_{\leq m}$ (in the augmented case). See e.g. [Tag 0183](http://stacks.math.columbia.edu/tag/0183) for a reference (treating the non-augmented case; the augmented case is similar). 


The point is that $(\Delta/[n])^{\operatorname{op}}\_{\leq n-1}$ is coinitial in $(\Delta\_a/[n])^{\operatorname{op}}\_{\leq n-1}$ when $n \geq 2$ (in the sense of [Tag 09WP](http://stacks.math.columbia.edu/tag/09WP); the definition on Wikipedia is too weak). Hence, you're computing the same limit by [Tag 002R](http://stacks.math.columbia.edu/tag/002R).


(The second condition in [Tag 09WP](http://stacks.math.columbia.edu/tag/09WP) boils down to checking $(\Delta/[n])^{\operatorname{op}}\_{\leq n-1}$ is still connected, since you're only removing the terminal object. Note also that this fails for $n=1$, and in this case you end up computing a product instead of a fibred product.)


