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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.

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Your phrasing of the definition is not quite right, or rather the word-ordering is not clear. In your initial sentence some of the quantifiers and ordering seem to be the wrong way round. Could you please check again that this is what you want to say? (For what it's worth, equalizers in many categories tend to have the flavour of being the "largest subobjects" on which two morphisms "agree", and are hence maximal rather than minimal in spirit. – Yemon Choi Jan 31 '10 at 10:55
Mh...yes, thank you. I translated from spanish and I missed something important: morphism $m$ should also satisfy $f\circ m=g\circ m$. Sorry. Anything new then? – Doctor Gibarian Jan 31 '10 at 12:01
it is still incorrect. $f e = g e$ cannot be provided for all $f,g$. rather, $f,g$ are fixed. – Martin Brandenburg Jan 31 '10 at 13:08
Thank you Martin - this is precisely what I was trying to get at. It only makes sense to speak of the equalizer of a parallel pair; in particular, the "for every" in the first sentence of the original post is incorrect – Yemon Choi Jan 31 '10 at 19:49
Thank you again (will I define the things properly one day?). – Doctor Gibarian Feb 1 '10 at 8:46
up vote 5 down vote accepted

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.

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