On the paradox that $n$-coskeletal simplicial sets model all homotopy types Please help me resolve the following paradox:


False claim: Let $X$ be an $n$-coskeletal, $n$-connected simplicial set. Then $X$ is weakly contractible.


Actually, I suppose the claim is true when $n=0$; but it is false for $n\geq 1$: nerves of posets realize all homotopy types, and these are 1-coskeletal.
Paradoxical "Proof" of false claim: Fix $m > n$. To show that $\pi_m(X) = 0$, consider a map $f: Sd^k \partial \Delta^{m+1} \to X$; we want to show that for some $l$, $Sd^{k+l} \partial \Delta^{m+1} \to Sd^{k} \partial \Delta^{m+1} \xrightarrow f X$ extends along $Sd^{k+l} \partial \Delta^{m+1} \to Sd^{k+l} \Delta^{m+1}$. 
Here $Sd$ is barycentric subdivision. Since $X$ is $n$-coskeletal, it suffices to show this after passing to $n$-skeleta. But then this problem can be solved because $X$ is $n$-connected, so that (after subdivision) lifts exist along any map between $n$-skeletal simplicial sets.
Resolution: The problem with the "proof" is that subdivision fails to commute with taking skeleta.
Paradoxical salvaged "proof": One can still try to build an extension of $f$ as follows. First, find an extension along $Sd^{l_1} sk_n Sd^k \partial \Delta^{m+1} \to Sd^{l_1} sk_n Sd^k \Delta^{m+1}$. To turn this into an extension along $sk_n Sd^{k+l_1} \partial \Delta^{m+1} \to sk_n Sd^{k+l_1} \Delta^{m+1}$ involves another lifting problem along a map between $n$-skeletal simplicial sets, so it can be solved after applying $Sd^{l_2}$. Continue in this manner, and you eventually construct an extension of $f$ along $Sd^\infty \partial \Delta^{m+1} \to Sd^\infty \Delta^{m+1}$. This shows that the homotopy group represented by $f$ is trivial.
Attempted resolution:
The process doesn't converge at a finite stage. The loophole must be that $Sd^\infty \partial \Delta^{m+1}$ doesn't have the homotopy type of $\partial \Delta^{m+1}$.
Questions:


*

*Is the map $\partial \Delta^{m+1} \to Sd^\infty \partial \Delta^{m+1} := \varinjlim_l Sd^l \partial \Delta^{m+1}$ a weak homotopy equivalence?

*If so, then how does one actually resolve the revived paradox?

*Even if not, is there a better way to formulate the resolution of the revived paradox? It feels as though a loophole was exploited.

*I think I'm starting to see why by the naure of "subdivision", no such operator is going to commute with taking skeleta. But if someone has a nice way to formulate why this is so, I'd love to hear it.
 A: As pointed out in the comments, the map in (1.) actually goes the other way around. Nevertheless, $l\colon\text{Sd}^{\infty}X\to X$ is a weak equivalence. Indeed, let $K$ be any Kan complex, $l$ induces a bijection between the sets of homotopy classes of maps 
\begin{equation*}
l^*\colon [X,K]\to [\text{Sd}^{\infty}X,K]\simeq [X,\text{Ex}^{\infty}K]\simeq [X,K]
\end{equation*} 
Where the first isomorphism comes from the fact that the adjunction $(\text{Sd},\text{Ex})$ is simplicial.
Now, I think the paradox in your proof is in the second sentence. I don't see why one should expect an extension along $\text{Sd}^l\text{sk}_n\text{Sd}^k\partial\Delta^{m+1}\to \text{Sd}^l\text{sk}_n\text{Sd}^k\Delta^{m+1}$ to exist (Unless this is somehow a consequence of the coskeletal condition). 
Let me explain why I don't think such an extension would exist in general.
We are trying to find a lift in the following diagram, for some $l$
$$\require{AMScd}
\begin{CD}
\text{Sd}^l\text{sk}_n\text{Sd}^k\partial\Delta^{m+1}@>f_{|\text{sk}_n}\circ \text{l.v.}>>X\\
@VVV\\
\text{Sd}^l\text{sk}_n\text{Sd}^k\Delta^{m+1}
\end{CD}
$$
where $\text{l.v.}$ is the last vertex map.
This is equivalent to finding a lift in the following diagram
$$\begin{CD}
\text{sk}_n\text{Sd}^k\partial\Delta^{m+1}@>j_X\circ f_{|\text{sk}_n}>>\text{Ex}^{\infty}X\\
@VVV\\
\text{sk}_n\text{Sd}^k\Delta^{m+1}
\end{CD}
$$
where $j_X\colon X\to \text{Ex}^{\infty}X$ is the usual inclusion.
But $\text{sk}_n\text{Sd}^k\partial\Delta^{m+1}\to\text{sk}_n\text{Sd}^k\Delta^{m+1}$ is not a weak-equivalence if $k\geq 1$, and so we have no reason to expect that such a lift exists.
To see why this map is not a weak equivalence, consider the fact that part of the $n$-skeleton of $\text{Sd}\Delta^{m+1}$ is contained in the "interior" of $\Delta^{m+1}$. 
This also gives you an example of (4), since if we reverse the order of subdivision and truncation, we get an isomorphism $\text{Sd}^k \text{sk}_n\partial\Delta^{m+1}\simeq \text{Sd}^k \text{sk}_n\Delta^{m+1}$.
