$\newcommand{\lan}{\operatorname{Lan}}\newcommand{\tr}{\operatorname{tr}}\newcommand{\H}{\mathsf{H}}\newcommand{\set}{\mathsf{Set}}\newcommand{\T}{\mathsf{Top}}\newcommand{\C}{\mathsf{C}}\newcommand{\psh}{\mathsf{Psh}}\newcommand{\op}{{^\mathsf{op}}}\newcommand{\sk}{\operatorname{sk}}\newcommand{\nat}{\mathsf{Nat}}$I don't know much about geometric realisation other than its very definition. The definition varies, but I prefer the 'clean' description in terms of Kan extension.
Notation: Let $\Delta$ be the simplicial category with objects indexed by the convention $n\sim\{0,1,2,\cdots,n\}$ for $n\in\Bbb N_0$. For any locally small category $\C$, let $\H:\C\to\psh_\C$ be the Yoneda embedding, with $\psh_\C$ the functor category $[\C\op,\set]$. Let $\Delta_\T:\H(\Delta)\to\T$ be the functor assigning to a standard $n$-simplex $\H(n)=\Delta(-,n)$ the standard affine $n$-simplex, and let it act on maps in the usual way. For any $n\in\Bbb N_0$, let $\Delta_{\le n}$ denote the full subcategory of $\Delta$ with objects $0,1,\cdots,n$, and let $\iota_n:\Delta_{\le n}\hookrightarrow\Delta$ be its inclusion functor. Let $\tr_n:\psh_\Delta\to\psh_{\Delta_{\le n}}$ be the $n$-truncation mapping $X\mapsto X\circ\iota_n$.
With that out of the way, we can describe the geometric realisation functor as follows: $$|\cdot|=\lan_{\H(\Delta)\hookrightarrow\psh_\Delta}(\Delta_\T):\psh_\Delta\to\T$$Where $\lan_\bullet$ is the left Kan extension. We can also define the $n$-geometric realisation (I 'made this up' myself, so it's possible this has a more standard name): $$|\cdot|_n:=\lan_{\H(\Delta_{\le n})\hookrightarrow\psh_{\Delta_{\le n}}}(\tr_n(\Delta_\T)):\psh_{\Delta_{\le n}}\to\T$$
My question: If $X$ is an $\le n$-dimensional simplicial set (that is, $X\cong\sk_n(X)$ – all $(m>n)$-simplices of $X$ are in the image of some $(k\le n)$-simplex) then is it true that: $$|X|\cong|\tr_n(X)|_n$$
Motivation: If this were true, it'd fit with my intuition and would make visualising simple examples of $|\cdot|$ much easier. For instance, to compute the geometric realisation of a two-dimensional simplicial set it'd suffice to examine only its triangles, edges and vertices (loosely speaking) – I wouldn't need to examine its higher order relations.
It is true that if $X'$ is another $\le n$-dimensional simplicial set with $\tr_n(X)\cong\tr_n(X')$ then $X\cong X'$. This is a simple corollary of the identity $\nat(\tr_nY,\tr_n Z)\cong\nat(\sk_n Y,Z)$, where the right-to-left direction of the isomorphism simply assigns $\varphi\mapsto\tr_n(\varphi)$. So, in particular, if two $\le n$-dimensional simplicial sets $X,X'$ share the same $n$-truncation, then $|X|\cong|X'|$.
This would suggest that the 'higher dimensional information' of $X$ is irrelevant and gets lost in the process of geometric realisation. This is almost obvious:
Stepping through the colimit formula for the left Kan extension, we find: $$|X|=\bigsqcup_{m=0}^\infty\bigsqcup_{\sigma\in X_m}|\Delta^m|/\sim$$Where $\sim$ is the smallest equivalence relation containing the identification: $$(x,X_f(\sigma))\sim(|f|(x),\sigma)$$For all $m,m'\in\Bbb N_0,\,f\in\Delta(m',m),\,\sigma\in X_m$ and $x\in|\Delta^{m'}|$.
As $X$ is $\le n$-dimensional, if $m>n$ and $\sigma\in X_m$, we have $\sigma=X_f(\tau)$ for some $\tau\in X_k$ and $k\le n$, $f\in\Delta(m,k)$ a surjection. Then for any $x\in|\Delta^m|$: $$(x,\sigma)\sim(|f|(x),\tau)$$And $|f|$ is also a surjection, so here the entire $m$-simplex $|\Delta^m|$ is collapsed onto $|\Delta^k|$. So it would seem that the higher dimensional simplices of $X$ do not contribute anything new to the colimit. By 'new', I mean the introducing of new points or a different topology as opposed to that found in $|\tr_n X|_n$.
However, there is one detail I can't figure out at the moment. Suppose $m,m'>n$ and $f\in\Delta(m,m')$, $\sigma,\sigma'\in X_m,X_{m'}$ have $X(f)(\sigma')=\sigma$. So there is an identification $(x,\sigma)\sim(|f|(x),\sigma')$. It is also true that $\sigma,\sigma'=X_{\alpha,\alpha'}(\tau,\tau')$ for some surjections $\alpha,\alpha'$ and $\le n$-dimensional simplices $\tau,\tau'$, so we have: $(x,\sigma)\sim(|\alpha|(x),\tau)$ and $(x,\sigma')\sim(|\alpha'|(x),\tau')$. Since an equivalence relation must be transitive, we deduce for all $x\in X_m$: $$(|\alpha|(x),\tau)\sim(|\alpha'f|(x),\tau')$$As a relation among the low-dimensional simplices. Perhaps this relation is lost in the quotient in $|\tr_n(X)|_n$ because the arrows $\alpha,\alpha'f$ are not present in $\Delta_{\le n}$. If so, then $|X|\cong|\tr_n(X)|_n$ would fail!
Can we prove that such relations are never lost? That is, is $|X|\cong|\tr_n(X)|_n$ always true?
Many thanks, and apologies in advance if this is trivial.
Best Answer
You're trying for an extremely manual argument, which is good for motivation, but not the most efficient for proving things. To flesh out Zhen's comment: a simplicial set is $n$-dimensional if and only if it's the left Kan extension of its restriction to the full subcategory $\Delta_{\le n}$ of $\Delta$ on objects no larger than $n.$ Your $n$-geometric realization is the left Kan extension of the geometric realization of $\le n$-dimensional simplices along Yoneda, so the point is that you can left Kan extend along $i_n:\Delta_{\le n}\to \Delta$ and then along the Yoneda embedding $y_\Delta$ of $\Delta,$ or just along $y_{\Delta_{\le n}},$ and you end up in naturally isomorphic places, since $y_\Delta\circ i_n=y_{\Delta_{\le n}}.$