Equalizer objects in Set.

Question

An equalizer in a category $\mathcal{C}$ is a couple $(E,eq)$ consisting in an object $E$ and a morphism $eq:E\longrightarrow X$ so that $f\circ eq=g\circ eq$ for every pair of parallel morphisms $f,g:X\longrightarrow Y$ and if for every other object $O$ and morphism $m:O→X$ there exists a unique morphism $u:O→E$ so that $eq\circ u=m$.

In the category $Set$, by taking sets and functions between them, an equalizer is the set of elements of the common domain where the functions are equal, that is: $Eq(f,g)=\{x\in X/f(x)=g(x)\}$ with $X$ a set and $(f,g)$ a couple of parallel morphism in Set.

My question is: can we say that the equalizer set is minimal among all the equalizer sets (like $O$ in the definition)?

Thanks for participate.

Answer 1

Given two parallel morphisms $f,g:X\to Y$ in some category $\mathcal{C}$, let us consider the category $\mathcal{E}_{f,g}$ :

Objects : all pairs $(E,e)$, where $E$ is an object of $\mathcal{C}$ and $e:E\to X$ is a morphism in $\mathcal{C}$ such that $f\circ e=g\circ e$,

Morphisms : from $(E',e')$ to $(E,e)$ are just morphisms $\varphi:E'\to E$ in $\mathcal{C}$ such that $e'=e\circ\varphi$.

Now an equaliser of $f,g$ is just a final object in the category $\mathcal{E}_{f,g}$. Final objects in any category are unique (provided they exist), up to a unique morphism; we may then talk of "the" equaliser of $f,g$.

When $\mathcal{C}$ is the category of sets, the equaliser always exists (and is therefore uniquely unique); as you say, it is the largest subset of the common source where the two maps coincide.

It is fine to think of it as a "maximal" object in $\mathcal{E}=\mathcal{E}_{f,g}$, but one must realise that it is also a "minimal" object in the opposite category $\mathcal{E}^\circ$ in the sense of being an initial object therein.