Given an adjunction $\langle F,G,\eta,\varepsilon\rangle\colon X\to A$, the canonical presentation of an object $a\in\operatorname{obj}(A)$ is the fork
(this is indeed a fork, by the naturality of $\varepsilon$).
In general, this fork need not be a coequalizer, but if $G$ is monadic, then we do get a coequalizer. To see this, note that the above pair is $G$-split: When applying $G$ to the fork, we get the split coequalizer
for the monad $\mathbb{T} := \langle T = G F,\eta,\mu=G\varepsilon F \rangle$ corresponding to the given adjunction and for the $\mathbb{T}$-algebra? $\langle x,h\rangle = \langle G a , G\varepsilon_a \rangle$. Hence, by the monadicity theorem, $G$ (in particular) reflects coequalizers for our pair.
The two parallel arrows $F G\varepsilon_a$ and $\varepsilon_{F G a}$ appearing in the canonical presentation have a common
section, namely, $F \eta_{G a}$ (by the triangle identities). Hence,
whenever $G$ is monadic, the resulting coequalizer is a reflexive coequalizer.
If $A = \mathbf{Grp}$, $X = \mathbf{Set}$ and $G$ is the forgetful functor, then for a group $a$, $F G a$ is just the free group on the elements of $a$, and $\varepsilon$ is the projection, taking a ‘’formal product’‘ of elements of $a$ to the actual product in $a$ (since by a triangle identity we have $G\varepsilon_a(\langle t\rangle)=t$ where $t\in Ga$ and $\langle t\rangle=\eta_{G a}(t)=$ the reduced word with one letter $t$).
Since the coequalizer of $\cdot\underoverset{f}{g}{\rightrightarrows}\cdot$ in $\mathbf{Grp}$ is the familiar quotient by the normal subgroup generated by elements like $f(t) g(t)^{-1}$, the canonical presentation (a coequalizer in this case) is indeed a presentation of $a$ in terms of generators and relations.
Categories Work, p. 153.
John Baez, ‘’Universal algebra and diagrammatic reasoning,’‘ slides available online
Last revised on February 20, 2012 at 21:51:03. See the history of this page for a list of all contributions to it.