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I want to understand if there is an intuition approchable with most possible 'elementary geometrical' knowledge for $n$-(co)skeleta of simplicial sets?

Formally sketleton & coskeleton functions arise as follows: For $\Delta$ the simplex category write $\Delta_{\leq n}$ for its full subcategory on the objects $[0],[1],\cdots,[n][0], [1], \cdots, [n]$. The inclusion $\Delta|_{\leq n} \hookrightarrow \Delta$ induces a truncation functor

$$\mathrm{tr}_n: \mathit{sSet}= [\Delta^{\mathrm{op}},Set] \to [\Delta_{\leq n}^{\mathrm{op}},\mathit{Set}]$$

that takes a simplicial set and restricts it to its degrees $\leq n$.

This functor has a left adjoint, given by left Kan extension $\mathrm{sk}_n: [\Delta_{\leq n},\mathit{Set}] \to \mathit{SSet}$ called the $n$-skeleton

and a right adjoint, given by right Kan extension $\mathrm{cosk}_n : [\Delta_{\leq n},Set] \to SSet$ called the $n$-coskeleton.

Now set $F: \Delta^{\mathrm{op}} \to Set, [n] \mapsto X_n$. The picture one conventionally has in mind thinking intuitionally/geometrically about $X$ is that one thinks $X_n$ as "the set of $n$-simplices/cells of the "simplicial complex" $X$ (only as geometrical intuition).

How can I think in this naive manner about $\mathrm{sk}_n(X)$ and $\mathrm{cosk}_n(X)$?

The $\mathrm{sk}_n(X)$ might be considered as a "subcomplex" of $X$ obtained from $X$ by killing all $m$-simplices with $m > n$. The way all $\ell$-simplices for $\ell \le n$ are "glued together" stays the same as for $X$, ie for $\ell$-simplices happens nothing.

If we keep thinking about $X$ as a simplicial complex, which picture should one have thinking about $\mathrm{cosk}_n(X)$? How it deviates from original $X$?

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For $k \le n$, the $k$-simplices in $\mathrm{cosk}_n(X)$ are the same as in $X$. For larger $k$, there is a unique $k$-simplex for every $n$-skeleton of a $k$-simplex you find in $X$, that is, $(\mathrm{cosk}_n(X))_k \cong \mathrm{Hom}(\mathrm{sk}_n \Delta^k, X)$.

You can also think inductively: again, for $k \le n$ the $k$-simplices in $\mathrm{cosk}_n(X)$ are the same as in $X$; then for each $k > n$ if you already know the simplices of dimension less than $k$ in $\mathrm{cosk}_n(X)$, you get the $k$-simplices by filling in uniquely every empty $k$-simplex you find in $\mathrm{cosk}_n(X)$. That is, for $k>n$, $(\mathrm{cosk}_n(X))_k \cong \mathrm{Hom}(\partial \Delta^k, \mathrm{cosk}_n(X))$.

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  • $\begingroup$ so maybe in more elementary language can we say $cosk_n(X)$ is constructed & glued in following way: The $k \le n$-simlices of $cosk_n(X)$ coinside with $X$. And for $k >n$ a $k$-simplex is contained in $(cosk_n(X))_k$ if all it's faces are contained in $cosk_n(X)_{k-1}$ and additionally are glued in compatible way to auch other? informally, if $cosk_n(X)$ already "contains" completly the boundary of this $k$-simplex, then it contains this $k$-simplex itself? Is this the correct intuitive "picture"? $\endgroup$ – MortyPB Jul 11 at 17:33
  • $\begingroup$ Yes, @MortyPB, that's the right idea. $\endgroup$ – Omar Antolín-Camarena Jul 11 at 17:37

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