# Visualize (co)sketeton of a simplicial set (geometrical intuition)

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 $$X$$?

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 $$k>n$$, $$(\mathrm{cosk}_n(X))_k \cong \mathrm{Hom}(\partial \Delta^k, \mathrm{cosk}_n(X))$$.
• 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"? – MortyPB Jul 11 '20 at 17:33