# How are the horizontal maps in the push out square defined? [closed]

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?

$$\newcommand{\sk}{\operatorname{sk}}\newcommand{\set}{\mathsf{Set}}\newcommand{\op}{{^{\mathsf{op}}}}\newcommand{\nd}{^{\mathsf{nd}}}$$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}
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.