Presentation of an algebra

Remarks

The presented algebra has this universal property. Let $B \in \mathcal{V}$ and let $f \colon X \to B$ be a map under which each relation $s = t$ holds in $B$. Then there is exactly one homomorphism $\langle X \mid R \rangle_{\mathcal{V}} \to B$ that extends $f$. Every $A \in \mathcal{V}$ has a presentation: take $X = A$ and $R$ the kernel of the homomorphism $F_{\mathcal{V}}(A) \to A$ that extends the identity map. Example: in the variety of commutative monoids, $\langle a \mid a^2 = a^3 \rangle$ is the monoid $\set{1, a, a^2}$ with $a^3 = a^2$. The relation holds there, and the universal property is checked by sending $a$ to any element $b$ with $b^2 = b^3$. Non-example: the same pair $(\set{a}, \set{a^2 = a^3})$ does not present $\Z / 2\Z$ as a commutative monoid, since in $\Z / 2\Z = \set{1, a}$ one has $a^2 = 1$ and $a^3 = a$, so the relation $a^2 = a^3$ fails. The presented algebra depends on the variety: $\langle a, b \mid \ \rangle$ is $\N^2$ in commutative monoids and the free monoid on two letters in monoids.