# What is an “absolute, equational pushout”?

I was leafing through Joyal and Tierney's Notes on simplicial homotopy theory.

In the first few lines of the section on the skeleton of a simplicial set they display the simplicial identities as a family of commutative squares involving the face and degeneracy maps. Right afterwards (on page 6, before Proposition 1.2.1) they state that "Now in such a diagram, the square of $\eta$’s is an absolute, equational pushout, and the square of $\varepsilon$’s is an absolute, equational pullback."

I was wondering what exactly this is supposed to mean. I know that an absolute colimit is a colimit that is preserved under all functors into another category, so I suppose this explains the "absolute" part of that quote.

Assuming $i \lt j$ and $n \geq 2$ their diagrams are $$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\qquad\large#1\qquad}\!\!\!\!\!\!\!\!\!} \newcommand{\la}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xleftarrow{\qquad\large#1\qquad}\!\!\!\!\!\!\!\!} \newcommand{\ua}[1] {\left\uparrow{\small#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\da}[1] {\left\downarrow{\small#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{cccccccccccccc} \!\!\!\![n] & \ra{\eta^j} & \!\!\!\![n-1] & & & & \!\!\!\![n] & \la{\varepsilon^{j}} & \!\!\!\![n-1] \\ \da{\eta^i} & & \da{\eta^i} & & & & \ua{\varepsilon^i} & & \ua{\varepsilon^i} \\ \!\!\!\![n-1] & \ra{\eta^{j-1}} & \!\!\!\![n-2] & & & & \!\!\!\! [n-1] & \la{\varepsilon^{j-1}} & \!\!\!\![n-2] \end{array}$$ and they state that the left hand diagram of (co)degeneracy maps is an "absolute, equational pushout" and the right hand diagram of (co)face maps is an "absolute, equational pullback".

So: What do they mean by "equational"?

Does this mean more than that the morphism provided by the universal property is given explicitly in terms of a map defined from the diagram? For example, if $f \colon [n-1] \to [p]$ and $g \colon [n-1] \to [p]$ are such that $f \eta^j = g \eta^{i}$ then the pushout map $h \colon [n-2] \to [p]$ is given by $h = \varepsilon^{i} f = \varepsilon^{j-1} g$.

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Thanks @Jack Schmidt for providing help on How to draw commutative diagrams! – Simplicial Presheaf Nov 22 '12 at 22:22
Anything that is a (co)limit for equational reasons will be an absolute (co)limit, but not all absolute (co)limits are "equational". The most well-known example of an "equational" colimit is the absolute coequaliser, as used in Beck's monadicity theorem. – Zhen Lin Nov 22 '12 at 22:51
@ZhenLin Thanks! If I understand correctly, you are confirming my suspicion that the "equational" nature of these pushout/pullback squares is basically explained by the proposition I linked to: given a square and maps satisfying a bunch of equations then it is an absolute pushout "just because the maps satisfy these equations", right? This sounds about what I was expecting. If you would be so kind as to post that as an answer I'd accept it. – Simplicial Presheaf Nov 22 '12 at 23:10
I haven't checked your example specifically, but you are free to post an answer to your own question if you have found a satisfactory proof on your own. – Zhen Lin Nov 22 '12 at 23:34