An **absolute pushout** is a pushout which is preserved by any functor whatsoever. In general this happens because the pushout is a pushout for purely “diagrammatic” reasons. See absolute colimit for more.

A particular pushout diagram in a particular category $C$ is an **absolute pushout** if it is preserved by every functor with domain $C$.

Equivalently, since the Yoneda embedding is the free cocompletion of $C$:

A particular pushout diagram in a particular category $C$ is an **absolute pushout** if it is preserved by the Yoneda embedding $C \hookrightarrow [C^{op},Set]$.

We propose the following notion of **split pushout**. See Isaacson, Def. 3.22.

A commutative square

defines a **split pushout** if there exist sections $p s = 1$, $q t = 1$, $m u = 1$

so that $p t = u n$.

Split pushouts are absolute pushouts.

Note that split pushouts are preserved by arbitrary functors, so it suffices to show that a split pushout is a pushout in the category in which it lives. To that end consider, a cone under the span $(p,q)$:

Upon composing with the sections to $q$ and $m$

we see that $c$ factors through the claimed pushout $P$ as $c = (b u) n$. We must verify that $b$ also factors as $b = (b u) m$. Since $p$ is an epimorphism, it suffices to prove that $b p = b u m p$, which follows easily:

$b p = c q = c q t q = b p t q = b u n q = b u m p.$

This produces the desired factorization. Finally, since $m$ is an epimorphism, such factorizations are unique.

Note the proof that a split pushout defines a pushout square in the category in which it lives did not require $p$ to be a *split* epimorphism. However, arbitrary functors do not preserve epimorphisms. They do however preserve split epimorphisms, and thus the section guarantees that the image of $p$ will define an epimorphism in any category.

A commutative square

is an absolute pushout if and only if either there exist

- A section $u:P\to B$, such that $m u = 1_P$.
- Morphisms $r_1,\dots,r_k : B \to A$ and $s_1,\dots,s_k : B\to A$, for some $k\ge 1$, such that $p s_1 = 1_B$, $q s_i = q r_i$ for all $i$, $p r_i = p s_{i+1}$ for all $i\lt k$, and $p r_k = u m$.
- Morphisms $t_1,\dots,t_{\ell+1} : C \to A$ and $v_1,\dots,v_{\ell} : C\to A$, for some $\ell \ge 0$, such that $q t_1 = 1_C$, $p t_i = p v_i$ for all $i\lt \ell$, $q v_i = q t_{i+1}$ for all $i\le \ell$, and $p t_{\ell+1} = u n$.

or the transpose thereof (i.e. interchanging $B$ with $C$ and so on).

For “if”, suppose given a commutative square

Since $m$ is a (split) epimorphism (by $u$), any factorization of this square through the given one will be unique, so it suffices to show that such a factorization exists. Define $x = b u:P\to X$. Then we have

$x m = b u m = b p r_k = c q r_k = c q s_k = b p s_k = b t r_{k-1} = \dots = b p s_1 = b$

and

$x n = b u n = b p t_\ell = c q t_\ell = c q v_{\ell-1} = b p v_{\ell-1} = b p t_{\ell-1} = \dots = c q t_1 = c.$

The transposed case is of course dual.

Conversely, suppose the given square is an absolute pushout. Thus, in particular the induced square

is a pushout in $Set$. Thus, in particular, the function $\hom(P,B)+\hom(P,C) \to \hom(P,P)$ is surjective, and thus for $1_P\in \hom(P,P)$ there must be either a $u:P\to B$ such that $m u = 1_P$ or a $u':P\to C$ such that $n u' = 1_P$. WLOG assume the former.

Now the induced square

is also a pushout in $Set$. We have two elements $1_B, u m \in \hom(B,B)$ that become equal in $\hom(B,P)$ (since $m 1_B = m = (1_P) m = m u m$), and in a pushout in $Set$ this means they must be related by a zigzag of elements of the vertex $\hom(B,A)$. Unraveling this explicitly produces the morphisms $r_i,s_i$.

Similarly, from the induced square

and the elements $1_C\in\hom(C,C)$ and $s n \in \hom(C,B)$, we obtain the morphisms $t_i,v_i$.

In particular, when $k=1$ and $\ell=0$, the above data reduces to

- A section $u:P\to B$, such that $m u = 1_P$.
- Morphisms $r,s : B \to A$ such that $p s = 1_B$, $q s = q r$, and $p r = u m$.
- A morphism $t : C \to A$ such that $q t = 1_C$ and $p t = u n$.

This is precisely the data of the above-defined notion of split pushout, together with the additional morphism $r:B\to A$ such that $q r = q s$ and $p r = u m$. However, given a split pushout as above we can define $r = t q s$ and check $q r = q t q s = q s$ and $p r = p t q s = u n q s = u m p s = u m$. This gives another proof that any split pushout is an absolute pushout.

In their study of generalized Reedy categories, Berger and Moerdijk introduce the notion of an Eilenberg-Zilber category, one of the axioms of which demands that spans of split epimorphisms admit absolute pushouts. In practice, this seems to be the case because the pushout of these split epimorphisms is a split epimorphism as above, often with an additional section $v$ of $n$ satisfying the additional equation that $v m = q s$.

The general characterization of absolute pushouts appears as Proposition 5.5 in:

- Robert Paré, Robert On absolute colimits. J. Algebra 19 (1971), 80–95.

The above notion of split pushout appears in Definition 3.22 of:

- Isaacson, Samuel B. “Symmetric cubical sets.” Journal of Pure and Applied Algebra 215.6 (2011): 1146-1173. (doi)

The Berger-Moerdijk definition of an Eilenberg-Zilber category appears in:

- Clemens Berger and Ieke Moerdijk,
*On an extension of the notion of Reedy category*(2008) (arXiv:0809.3341)

Last revised on December 8, 2022 at 20:28:35. See the history of this page for a list of all contributions to it.