Equalizers & Coequalizers

Cut a domain down to where two maps agree, or squash a codomain until they must — and discover that kernels and quotient groups were these all along.

The idea

The equalizer and the coequalizer are the two universal constructions attached to a parallel pair of arrows $f, g \colon A \to B$. The equalizer cuts $A$ down to the part on which $f$ and $g$ already agree; the coequalizer collapses $B$ until they are forced to agree. Neither is defined by a formula: each is defined by a universal property, and the formula is derived afterwards in each particular category.

Definition (Equalizer).

Fix a parallel pair of arrows $f, g \colon A \to B$. A fork on the pair is an arrow $q \colon T \to A$ with $f \circ q = g \circ q$. An equalizer of $f$ and $g$ is a fork $e \colon E \to A$ that every fork factors through exactly once: for each fork $q \colon T \to A$ there is exactly one arrow $\bar{q} \colon T \to E$ with $e \circ \bar{q} = q$.

Reversing every arrow gives the coequalizer.

To see the universal property choose an answer, work in $\mathbf{Set}$. Take $E = \{a \in A : f(a) = g(a)\}$ with its inclusion $e$ into $A$. A fork $q$ satisfies $f(q(t)) = g(q(t))$ for every $t$, so its values already lie in $E$; reading $q$ as a map into $E$ gives the factorisation, and no other arrow works, since $e \circ \bar{q} = q$ forces $\bar{q}$ to take the same values as $q$. Dually, the coequalizer is the quotient of $B$ by the smallest equivalence relation making $f(a)$ equivalent to $g(a)$.

Two constructions met separately in algebra are instances: a kernel is the equalizer of a homomorphism against the trivial map, and a quotient group is a coequalizer.

Ways to work on it

• Walkthrough. The equalizer of a parallel pair and its dual, computed in Set, Grp, Ab and Top, with kernels and quotients falling out as special cases.
• Proof. Why every equalizer is a monomorphism — the argument in which the uniqueness half of a universal property does all the work.
• Practice. Compute an equalizer in Set, count the classes of a coequalizer, and read a kernel off as an equalizer against the trivial homomorphism.
• Hardest. Candidates that fail, a coequalizer in Ab that has to be computed, and a monomorphism that is no equalizer.

Not sure where to start? Take the ten-question placement test.