$\newcommand{\lan}{\operatorname{Lan}}\newcommand{\tr}{\operatorname{tr}}\newcommand{\H}{\mathsf{H}}\newcommand{\set}{\mathsf{Set}}\newcommand{\T}{\mathsf{Top}}\newcommand{\C}{\mathsf{C}}\newcommand{\psh}{\mathsf{Psh}}\newcommand{\op}{{^\mathsf{op}}}\newcommand{\sk}{\operatorname{sk}}\newcommand{\nat}{\mathsf{Nat}}$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.