split coequalizer

For purposes of this page, a fork (some might say a “cofork”) in a category $C$ is a diagram of the form

$A \;\underoverset{f}{g}{\rightrightarrows}\; B \overset{e}{\rightarrow} C$

such that $e f = e g$. A **split coequalizer** is a fork together with morphisms $s\colon C\to B$ and $t\colon B\to A$ such that $e s = 1_C$, $s e = g t$, and $f t = 1_B$. This is equivalent to saying that the morphism $(f,e)\colon g \to e$ has a section in the arrow category of $C$.

The name “split coequalizer” is appropriate, because in any split coequalizer diagram, the morphism $e$ is necessarily a coequalizer of $f$ and $g$. For given any $h\colon B\to D$ such that $h f = h g$, the composite $h s$ provides a factorization of $h$ through $e$, since $h s e = h g t = h f t = h$, and such a factorization is unique since $e$ is (split) epic. In fact, a split coequalizer is not just a coequalizer but an absolute coequalizer: one preserved by all functors.

On the other hand, suppose we are given only $f,g\colon A\to B$ and $t\colon B\to A$ such that $f t = 1_B$ and $g t f = g t g$ (which is certainly the case in any split coequalizer, since $g t f = s e f = s e g = g t g$). Such a situation is sometimes called a **contractible pair**. In this case, any coequalizer of $f$ and $g$ is split, for if $e\colon B\to C$ is a coequalizer of $f$ and $g$, then the equation $g t f = g t g$ implies, by the universal property of $e$, a unique morphism $s\colon C\to B$ such that $s e = g t$, whence $e s e = e g t = e f t = e$ and so $e s = 1_C$ since $e$ is epic.

Similarly, if $e\colon B\to C$ splits the idempotent $g t$ with section $s\colon C\to B$, so that $e s = 1$ and $s e = g t$, then we have

$e g = e s e g = e g t g = e g t f = e s e f = e f$

and the other identities are obvious; thus $e$ is a split coequalizer of $f$ and $g$.

Dually, if $e\colon B\to C$ is a split epimorphism, with a splitting $s\colon C\to B$, say, then $e$ is a split coequalizer of $B \;\underoverset{1}{s e}{\rightrightarrows}\; B$, the morphism $t$ being the identity.

Moreover, $e$ is also the split coequalizer of its kernel pair, if the latter exists. For if $A \;\underoverset{f}{g}{\rightrightarrows}\; B$ is this kernel pair, then the two maps $s e, 1_B \colon B\to B$ satisfy $e \circ s e = e \circ 1_B$, and hence induce a map $t\colon B\to A$ such that $f t = 1_B$ and $g t = s e$.

The “ur-example” of a split coequalizer is the following. Let $A$ be an algebra for the monad $T$ on the category $C$, with structure map $a\colon T A \to A$. Then the diagram

$T^2 A \;\underoverset{\mu_A }{T a}{\rightrightarrows}\; T A \overset{a}{\rightarrow} A\, ,$

called the canonical presentation of the algebra $(A,a)$, is a split coequalizer in $C$ (and a reflexive coequalizer in the category of $T$-algebras). The splittings are given by $s = \eta_A \colon A \to T A$ and $t = \eta_{T A} \colon T A \to T^2 A$. (Here $\mu$ and $\eta$ are the multiplication and unit of the monad $T$.)

This split coequalizer figures prominently in Beck’s monadicity theorem.

Revised on January 10, 2012 18:23:15
by Finn Lawler
(86.41.35.83)