In the answer given by @Kevin Arlin in the MSE question https://math.stackexchange.com/a/2994934/820022 if I am not mistaken the geometric realization of a simplicial set $X$ is defined as a colimit of the following diagram in Top :
$\pi \circ p:\Delta \downarrow X \rightarrow Top$ where $\pi:\Delta \rightarrow Top$ is defined as the canonical functor sending $[n] \rightarrow |\Delta^n|$ in object level (where $|\Delta^n|$ is the standard geometric $n$-simpplex) and also appropriately defined in the morphism level whereas I guess $p:\Delta \downarrow X \rightarrow \Delta$ is defined as follows:
On Objects: $(\sigma:\Delta^n \rightarrow X) \mapsto [n] \in \Delta,$ the usual finite ordinal category.
On Morphisms: 
$\theta \mapsto (\theta_{*}:[n] \rightarrow [m])$
where $\theta_{*}:[n] \rightarrow [m]$ is defined as $\theta_{[n]}(1_{[n]})$ (Coming from contravariant Yoneda lemma).
I was trying to show that $p$ is indeed a functor but I am struck at the following step while showing $(\psi \circ \phi)_{*}= \psi_{*} \circ \phi_{*}$ where $\psi:\Delta^m \rightarrow \Delta^{r}$ and $\phi: \Delta^{n} \rightarrow \Delta^{m}$. (Though I made abuse of notation for convenience).
My confusion:
$\psi_{*} \circ \phi_{*}= \psi_{[m]}(1_{[m]}) \circ \phi_{[n]}(1_{[n]})$......(1)
$(\psi \circ \phi)_{*}=(\psi \circ \phi)_{[n]}(1_{[n]})= \psi_{[n]} \circ \phi_{[n]}(1_{[n]})$.....(2)
I am not able to show (1) = (2).
Am I misunderstanding anything?
Or did I interpret the answer by @Kevin Arlin in an incorrect way?
Thanks in advance.
Let $Y:\Delta \rightarrow sSets$ be the Yoneda embedding. By Yoneda embedding there exist $g:[m] \rightarrow [r]$ and $f:[n] \rightarrow [m]$ such that $\psi = Y(g)$ and $\phi= Y(f)$.
So $\psi \circ \phi=Y(g) \circ Y(f)$.
Now component wise, $(\psi \circ \phi)_{[k]}= \psi_{[k]} \circ \phi_{[k]}= Y(g)_{[k]} \circ Y(f)_{[k]}$ for $[k] \in \Delta$.
So, in particular $\psi_{[m]}(1_{[m]}) \circ \phi_{[n]}(1_{[n]})=Y(g)_{[m]}(1_{[m]}) \circ Y(f)_{[n]}(1_{[n]})= g \circ 1_{[m]} \circ f \circ 1_{[n]}=g \circ f= \psi_{*} \circ \phi_{*}$ (using the same notation as mentioned in the question.)
On the other hand, $(\psi \circ \phi)_{*}= (\psi \circ \phi)_{[n]}(1_{[n]})= (Y(g) \circ Y(f))_{[n]}(1_{[n]})=Y(g \circ f)_{[n]}(1_{[n]})= g \circ f \circ 1_{[n]}= g \circ f$.
Hence $\psi_{*} \circ \phi_{*}= (\psi \circ \phi)_{*}$.
So, $p:\Delta \downarrow X \rightarrow \Delta$ is a functor. (Proved)
(Indentity Preservation is easy to show.)