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Consider the inclusion $i_n: \varDelta_{\leq n} \hookrightarrow \varDelta$. This induces a truncation functor $tr_n:\mathbf{sSet}\to\mathbf{sSet}_{\leq n}$, which has a left and a right adjoint, $tr_n^L$ and $tr_n^R$, obtained via Kan extension. This allows us to define the $n$-(co)skeleton of a simplical set $X$: $$ sk_n X := tr_n^L tr_n X, \quad cosk_n X := tr_n^Rtr_n X. $$ It is easy to see that these functors form an adjunction.

$sk_n$ is analogous to forgetting about cells of dimension greater than $n$ in a CW complex, i.e. taking the n-skeleton. What is the dual analogy (if it exists) for the coskeleton of a simplical set $X$?

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    $\begingroup$ You glue higher dimensional cells along each boundary for dimension greater than $n$. When your simplicial set is a Kan complex, this is equivalent to killing higher homotopy groups (again greater than $n$). $\endgroup$
    – user40276
    Commented Jun 28, 2016 at 22:52
  • $\begingroup$ Also check out the “Formal definition” section of en.wikipedia.org/wiki/Hypercovering. Except the third paragraph is probably slightly wrong as one need to include the augmentation map into the diagram of which the limit is taken. Also I suspect that cosk sk isn’t necessary; just cosk will do. $\endgroup$
    – Hypatia du Bois-Marie
    Commented Nov 4, 2022 at 7:53

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It should be added that if $X$ is a Kan complex, then $X\to \mathrm{cosk}_n X$ computes a model for the $(n-1)$-truncation of the homotopy type of $X$. That's a fact you can easily read off using elementary definition of homotopy groups of a Kan complex, together with the bijection $$ \{ \Delta^k \to \mathrm{cosk}_n X \} \Leftrightarrow \{ \mathrm{sk}_n \Delta^k \to X\}. $$

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A simplicial set $X$ is $k$-coskeletal iff the following condition holds:

a simplex of dimension $\geq k$ is present iff all of its $(k-1)$-dimensional faces are present in $X$.

A standard example is the usual nerve of a small category, which is readily seen to be $2$-coskeletal. (Similarly, the nerve of an $n$-category is $(n+1)$-coskeletal).

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    $\begingroup$ Your last point is not correct: the nerve of a category of 2-coskeletal, and the nerve of an n-category should be (n+1)-coskeletal. See eg ncatlab.org/nlab/show/simplicial+skeleton#examples and references there. $\endgroup$
    – David Roberts ♦
    Commented Aug 16, 2018 at 7:28
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    $\begingroup$ Phrasing @David's comment as a fuzzy slogan: In a category, not every triangle commutes, but every tetrahedron (or higher structure) of morphisms made from commutative triangles commutes. $\endgroup$
    – Ingo Blechschmidt
    Commented Nov 3, 2018 at 21:37
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    $\begingroup$ Thanks @davidroberts, now fixed I hope. $\endgroup$
    – Vidit Nanda
    Commented Nov 11, 2019 at 16:19
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For a simplicial set $X$ you can see the $n$-th skeleton to have no $m$-simplices other than the degenerate ones for $m>n$.(As you say)

For $m\leq n$ both the $n$-skeleton and the $n$-coskeleton have the same $m$-simplices as $X$.

And to answer your question for $m>n$ the $n$-coskeleton has $m$-simplices ALL possible configurations of a $m+1$ number of $(m-1)$-simplices in $\mathrm{cosk}_nX$ that agree in the appropriate $m-2$-faces to form the shape of a standard m-simplex.

If I say it in a lower dimension it will sound less tongue-twisty and easier to imagine it geometrically, which I think is what you are looking for: Take $n=2$

  • For $m=3$ the 3-simplices of the 2-coskeleton are all four possible triangles in $X$ that you can arrange in a tetrahedral shape.

  • For $m=4$ the 4-simplices are all five possible tetrahedra in $\mathrm{cosk}_2X$ that form the shape of a standard 4-simplex (pentachoron).

So in this sense, the coskeleton is the largest simplicial set containing X and the skeleton the smallest.

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  • $\begingroup$ Perhaps I would rephrase the first dot point as: For $m=3$ the 3-simplices of the 2-coskeleton correspond to all possible choices of four triangles in $X$ which can be arranged into a tetrahedral shape.$^\dagger$ and finish at the end with: $\dagger$ I.e. to a map on the $2$-truncations $(\Delta^3)^{(2)}\to X_\bullet^{(2)}$, which is determined by where the $4$ non-degenerate $2$-simplicies are mapped, such that these form a tetrahedron. $\endgroup$
    – Skies burn
    Commented Feb 25, 2020 at 0:22

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