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 $cosk_{n-1}(sk_{n-1}(U_*))$.

[DI] Dugger-Isaksen "Hypercovers in topology"


1 Answer 1


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 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; the definition on Wikipedia is too weak). Hence, you're computing the same limit by Tag 002R.

(The second condition in 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.)

  • $\begingroup$ Ah, this is very helpful. Thank you! $\endgroup$
    – Anonymous
    Dec 29, 2016 at 14:46

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