I'm currently trying to understand Lemma 14.2 of Peter May's Simplicial Objects in Algebraic Topology, which (paraphrasing) states that given a simplicial set $X$ and a point in the coproduct $(x,p) \in \coprod_{n \geq 0} X_n \times |\Delta^n|$, there is a unique non-degenerate pair $(y,q)$ equivalent to $(x,p)$, where $\sim$ is the equivalence relation by which we quotient to define the geometric realization of $X$. Concretely, it is generated by
$$ (Xd(x),p) \sim (x,|d|(p)) \text{ and } (Xs(x), p) \sim (x,|s|(p)) \tag{1} $$
for each coface map $d$ and codegeneracy map $s$ in $\Delta$.
Recall also that a pair $(y,q)$ is non-degenerate if $y$ is non degenerate, i.e. there is no simplex $z$ and degeneracy map $s$ such that $y = s(z)$, and $p$ is interior, that is , there is no coface map $d$ such that $p = |d|(q)$ for some $q$.
I have already proved that for any simplex $x$ there is a unique non degenerate $y$ and a composition $s$ of degeneracy maps such that $x = s(y)$, and similarly there is a unique point $q$ on a topological standard simplex and a composition of coface maps $d$ such that $p = |d|(q)$.
Now, the proof of the lemma proceeds as follows: by the last paragraph, if $x = s(q)$ with $s = X\sigma$ and $p = d(q)$ are the previous decompositions, we can define $\lambda(x,p) := (y,\sigma (p))$ and $\rho(x,p) := (Xd (x), q)$.
These mappings satisfy $\lambda(x,p) \sim (x,p) \sim \rho(x,p)$ and so does their compostion $\lambda \rho$, whose image is comprised uniquely of non-degenerate pairs. This shows existence.
The proof ends by claiming that $\lambda \rho$ sends equivalent points to the same non-degenerate pair (since $\lambda \rho$ fixes the latter, this proves uniqueness). It is not clear to me why this is the case, although I see that it suffices to prove it only for the equivalent pairs in $(1)$.
It holds for pairs 'related by a coface map' since given $(x,p)$ with $p=|d|(q)$ and $q$ interior we have
$$ \begin{align} \rho(Xd^i(x),x) &= \rho(Xd^i(x),|d|q) = (XdXd^i(x),q)\\ &= (X(d^id)(x),q) = \rho(x,|d^id|(q)) = \rho(x,|d^i|p). \end{align} $$
However, for pairs $(x,p) \sim (x',p')$ 'related by a degeneracy map', it is clear to me that $\lambda(x,p) = \lambda(x',p')$ by the same argument, but this does not say that $\lambda \rho(x,p) = \lambda \rho(x',p')$.
I tried to make a direct computation to show that first composing with $\rho$ does not break things, but I haven't managed to conclude. Is there any (preferably painless) way to show this?
Any help is greatly appreciated.
Edit: I have an additional question. The image of $\lambda \rho$ consists only of non degenerate points because 'collapses' of interior points are interior, right? I think I have managed to prove this but strongly using that I was working with topological simplices (specifically, I showed that a point $a_1e_1 + \dots + a_ne_n \in |\Delta^n|$ has non-vanishing coordinates if and only if it is interior, and a collapse $s^i$ preserves this property).
(I'm using $d$, $s$, and $t$ for maps in $\Delta^{\text{op}}$ and $|\cdot|$ for the corresponding maps in $\Delta$.)
Given $(sx,p)$, let $p=|d|q$ where $q$ is interior, let $ds=s'd'$, and let $d'x=tz$ where $z$ is nondegenerate. Note that $|s|p=|s||d|q=|d'||s'|q$, where $|s'|q$ is interior. Then we have
$\lambda \rho (sx,p) = \lambda(d(sx),q) = \lambda(s'(d'x),q)=\lambda(s'tz,q)= (z,|s't|q)=(z,|t||s'|q)$
and
$\lambda \rho (x,|s|p) = \lambda(d'x,|s'|q)=\lambda(tz,|s'|q)=(z,|t||s'|q)$,
so $(sx,p)$ and $(x,|s|p)$ are indeed sent to the same thing by $\lambda\rho$.
And the answer to your second question is "yes"; it works because collapses of interior points are interior.