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 $X$?



Visualize (co)sketeton of a simplicial set (geometrical intuition)
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 $k>n$, $(\mathrm{cosk}\_n(X))\_k \cong \mathrm{Hom}(\partial \Delta^k, \mathrm{cosk}\_n(X))$.


