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

$F G F Ga\underoverset{F G\varepsilon_a}{\varepsilon_{F G a}}{\rightrightarrows}F G a
\overset{\varepsilon_a}{\rightarrow}a$

(this is indeed a fork, by the naturality of $\varepsilon$).

Properties

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.

Example

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.