Every model of an algebraic theory is a quotient of a free model I have stumbled upon the following proposition from Borceux:

Proposition 3.8.9 Let $\mathcal{T}$ be an algebraic theory. Every $\mathcal{T}$-model $M$ is a quotient of a free model. More precisely, the following diagram is a coequalizer:
  $$FUFU(M) \overset{\varepsilon_{FU(M)}}{\underset{FU(\varepsilon_M)}{\rightrightarrows}} FU(M) \xrightarrow{\varepsilon_M} M.$$

Here $\varepsilon$ is the counit the free-forgetful functor adjunction.
In contexts with kernels I'd think the kernel of $\varepsilon _M$ would simply be the relations that hold in $M$. But I'm not sure how to concretely make sense of these coequalizer diagram. What's the information in it in terms of generators and relations? Can someone break it down for me?
 A: Looking at examples is probably very helpful here. Take as your algebraic theory the theory of commutative algebras. Let $M$ be some commutative algebra. $FU(M) = S(M)$ is the free symmetric (polynomial) algebra generated by the vector space $M$, and $FUFU(M) = S(S(M))$ is the polynomial algebra on that. Elements of $S(M)$ are "parenthesized" formal products of elements of $M$, e.g. $(a \wedge b \wedge b) \wedge (c \wedge c \wedge c \wedge d) \in S(S(M))$ is a polynomial of weight $2$, the product of $a \wedge b \wedge b$ and $c \wedge c \wedge c \wedge d$ (both of which are in $S(M) \hookrightarrow S(S(M))$).
The map $\varepsilon_{FU(M)}$ removes the outer parentheses, and the element I wrote above is sent to $a \wedge b \wedge b \wedge c \wedge c \wedge c \wedge d \in S(M)$. On the other hand, $FU(\varepsilon_M)$ performs the product in $M$ in all the inner expressions, so the element above is sent to $ab^2 \wedge c^3 d \in S(M)$. Obviously both of these morphisms are equalized by $\varepsilon_M$, and the final result is $ab^2c^3d \in M$.
So concretely, to summarize, this says that $M$ has a presentation where generators are elements of $M$, and relations says that if you perform an operation from your algebraic theory on the generators (elements of $M$), then you get the generator associated to the element of $M$ obtained by actually performing the operation in $M$.
It's actually a standard trick used in group theory to prove that every group has a presentation: for a group $G$, consider the free group $F(G)$ on the set $g$, generated say by $\{x_g\}_{g \in G}$, and take as relations $x_g x_h \equiv x_{gh}$. Then this gives a presentation of $G$. The proposition you've quoted is a generalization of that fact.
