I'm on my way reading this article of Milnor about the geometric realisation of a "(complete) semi-simplicial complex" ( = simplicial set, in modern terms).
I encountered a problem in a passage of the proof of theorem 2 page 359 which is the well-known results that for simplicial sets $X,Y$
$$|X\times Y|\cong |X|\times |Y|$$
I found the very same proof in a lot of other references about this well-known result, there is the very same proof (see e.g. May's The Geometry of Iterated Loop Spaces) and therefore the same passage bother's me.
Minor defined an inverse for $\eta:=|\rho_1|\times |\rho_2|$ (the two projections) and call it $\bar{\eta}$, $$\eta\circ \bar{\eta}=Id$$ is trivially seen to be true.
I can't see why $$\bar{\eta}\circ\eta =Id$$
It should be immediate (due to the lack of explanations). What's unclear is that in the last equality they directly write $\delta_n$ as second component after applying $\bar{\eta}$ to $(|k_n, \delta_n| , |k_n',\delta_n|)$ where in the definition of $\bar{\eta}$ it's clear that we are enlarging the dimension of the simplex, (i think we are double it) and I can't see how we can retrieve this case. Reading more carefully this should be a consequence of the fact we can take non-degenerate simplices, but I can't see how to use this hypothesis.
I hope my doubt is clear, what I'm asking is why the second equality in the proof of $\bar{\eta}\circ \eta =Id$ is true.
First note that for $(\lambda_0,\dots,\lambda_{n+1}) := \delta_n \in \Delta^n$, we have $0 = \lambda_0 < \dots < \lambda_{n+1} = 1$, because $\delta_n$ is interior.
To calculate $$\bar{\eta}(|k_{n-p},s_{i_1} \dots s_{i_p} \delta_n|,|k'_{n-q},s_{j_1} \dots s_{j_q} \delta_n|)$$ via the definition of $\bar{\eta}$, observe that $s_{i_1} \dots s_{i_p} \delta_n$ arises from $\delta_n = (\lambda_0,\dots,\lambda_{n+1})$ by omitting those components with index $i \in \{i_1+1,\dots,i_p+1\}$, and that $s_{j_1} \dots s_{j_q} \delta_n$ arises from $\delta_n = (\lambda_0,\dots,\lambda_{n+1})$ by omitting those components with index $j \in \{j_1+1,\dots,j_q+1\}$. Because "the indices $i_{\alpha}$ and $j_{\beta}$ must be distinct", if we collect all the distinct numbers from $s_{i_1} \dots s_{i_p} \delta_n$ and $s_{j_1} \dots s_{j_q} \delta_n$ (and arrange them in order), we do indeed get back $(\lambda_0,\dots,\lambda_{n+1})$, so that (in Milnor's notation) $\delta_n'' = \delta_n$.
I hope this will give you enough impetus to comprehend the rest of the proof, too.