How are the horizontal maps in the push out square defined?

Question

Let A be an Eilenberg-Zilber category and $X\subset Y$ be presheaves over A. For any non-negative integer n, there is a canonical push out square $\require{AMScd}$ \begin{CD} \sqcup_{y\in \sum}\partial h_{a} @>{}>> X \cup Sk_{n-1}(Y)\\ @VVV @VVV\\ \sqcup_{y\in \sum} h_{a} @>{}>> X \cup Sk_{n}(Y) \end{CD} where $\sum$ denotes the set of non-degenerate sections of $Y$ over $a$ that do not belong to $X$, and such that d($a$) = n. What are the horizontal morphisms in the above square? How are they defined?

Answer 1

$\newcommand{\sk}{\operatorname{sk}}\newcommand{\set}{\mathsf{Set}}\newcommand{\op}{{^{\mathsf{op}}}}\newcommand{\nd}{^{\mathsf{nd}}}$I only know about simplicial sets, but hopefully the answer for that can help you in general. That is, I work in $A=[\Delta\op,\set]$ where $\Delta$ is the simplicial category. For any $n\in\Bbb N_0$ let $\Delta^n$ and $\partial\Delta^n$ be the $n$th representable simplicial set and its boundary, respectively.

In this instance we set, for all $n\in\Bbb N_0$, $\Sigma_n:=Y_n\nd\setminus X_n\nd$ and we want a pushout: $$\begin{align}&\bigsqcup_{\sigma\in\Sigma_n}\partial\Delta^n\quad\quad\overset{f}{\longrightarrow}&X\cup\sk_{n-1}Y\\&\,\,\,\iota\downarrow&\downarrow\iota\,\,\,\,\,\,\,\,\,\,\\&\bigsqcup_{\sigma\in\Sigma_n}\Delta^n\quad\quad\overset{g}{\longrightarrow}&X\cup\sk_nY\end{align}$$

(that formats nicely on my computer but may break on other devices, let me know if there is a problem)

This is a case of the maps being "obvious" and the reader has to struggle past the author's terseness. By the Yoneda lemma, every $\sigma\in\Sigma_n$ yields some $\overline{\sigma}:\Delta^n\to Y$, and this restricts to a map $\partial\Delta^n\to\sk_{n-1}Y$ and we get a map into $X\cup\sk_{n-1}Y$ in this way. $f$ is simply what you get when you apply the universal property of coproducts to this collection of arrows $\partial\Delta^n\to\sk_{n-1}Y$.

$g$ is defined very similarly, for each $\overline{\sigma}:\Delta^n\to Y$ corestricts to some $\overline{\sigma}:\Delta^n\to\sk_n Y\hookrightarrow X\cup\sk_n Y$, and these arrows assemble to an arrow $g$ out of the coproduct. Verifying the pushout is indeed a pushout was not as super obvious to me as my own textbooks thought it was.