On proving that the geometric realization of $\Delta^n = \Delta(-,n)$ is homeomorphic to $|\Delta^n|$. I'm trying to prove that the geometric realization of $\Delta^n = \Delta(-,n) : \Delta^{op} \to \mathsf{Set}$ coincides with the geometric realization of $\Delta^n$ as a simplicial complex (I will note $|\Delta^n|$ only for the latter, to avoid confusion).
I'm working with the definition of geometric realization given by
$$
|X| = \left(\coprod_{k \geq 0} X_k \times |\Delta^k|\right) \Big/ \sim \tag{1}
$$
with each $X_k$ discrete, identifying $$(x,|d^i|(p)) \sim (d_i(x),p) \text{ and }(x,|s^i|(p)) \sim (s_i(x),p),$$with $d_i,s_i$ the face and degeneracy maps of the simplicial set $X$ and $d^i, s^i$ the coface and codegeracy maps of the standard simplices. 
My idea was to prove that $\{id_n\} \times |\Delta^n|$ is a fundamental domain for $\sim$, since it is both compact and homeomorphic to $|\Delta^n|$.
Given a point $(f,q)$, we know that $f : k \to n$ can be written as
$$
f = f_1 \circ \cdots \circ f_n
$$
with $f_i = s^{j_i}$ or $f_i = d^{j_i}$ being some coface/codegeneracy maps. Since the face and degeneracy maps on $\Delta^n$ are given by the precomposition of coface and codegeneracy maps, we have that
$$
\begin{align}
(f,q) = (f_n^* \cdots f_1^* (id),q) = (id, |f_1 \cdots f_n|(q)) \in \{id\} \times |\Delta^n|.
\end{align}
$$ 
This shows that each point of the geometric realization has a representative in $\{id\} \times |\Delta^n|$. 
Now, it is intuitive to me that the relation by which we divide in $(1)$ does not identify points between simplices of $X$ of the same dimension. Is this the case? If true, how can this be shown?
 A: Just observe that there is a map $|X|\to |\Delta^n|$: each element of $X_k=\Delta(k,n)$ determines a map $|\Delta^k|\to|\Delta^n|$, and so these maps together over all values of $k$ give a map $\coprod_kX_k\times|\Delta^k|\to|\Delta^n|$.  This map respects the equivalence relation $\sim$, essentially by definition (if two elements of $X_k\times|\Delta^k|$ and $X_{k+1}\times|\Delta^{k+1}|$ are related by a face or degeneracy, the corresponding maps $|\Delta^k|\to |\Delta^n|$ and $|\Delta^{k+1}|\to|\Delta^n|$ are related by composing with a map between $|\Delta^k|$ and $|\Delta^{k+1}|$ and this gives exactly the relation required by $\sim$), and thus descends to a map $f:|X|\to |\Delta^n|$.
If $i:|\Delta^n|\to |X|$ is the map given by the inclusion of $\{id_n\}\times|\Delta^n|$, then it is clear that $fi$ is the identity, and so in particular $i$ is injective.  On the other hand, the work you have done proves exactly that $i$ is surjective.  Thus $i$ is a bijection, and $f$ is its inverse, and so they are inverse homeomorphisms.
