# Universal Arrow for coequalizers

This is for Sets and functions(total). For Objects $X,Y$ and two arrows $f,g: X \to Y$ a coequalizer should be $Y \to Q$ such that $q \circ f = q \circ g$ and given $g_2: Y \to Q_2$ there must be $u: Q \to Q_2$. Therefore $Q$ must be a set with the sets of equivalence relations. My question is what operation would the arrow $u: Q \to Q_2$ represent? Since $g_2: Y \to Q_2$ it is also a coequalizer arrow it seems that it does the same thing as $Y \to Q$. If yes then what distinguishes $Q$ from $Q_2$?

-
What distinguishes them is that there is a canonical arrow $u : Q \to Q_2$ but none in the opposite direction. – Adeel Feb 14 '14 at 17:34
So $Y \to Q$ and $u:Q \to Q_2$ are essentially the same arrows in terms of what they do? Are set objects $Q$ and $Q_2$ the same sets of sets besides the extra elements of each that are not mapped from the arrows? – user128828 Feb 14 '14 at 17:37
I guess you could say that $Y \to Q$ and $Y \to Q_2$ are "essentially the same" in the sense that the latter factors through the former. It may be helpful to consider an explicit example of an arrow $Y \to Q_2$. – Adeel Feb 14 '14 at 17:41

This definition of coequalizer has a missing part. Say $f,g$ and $q$ their coeqalizer, as defined in the question. Then, if there is $g_2:Y\to Q_2$, there must also be a unique $u:Q\to Q_2$ such, that $g_2=u\circ q$. Not just an arbitrary $u$.
$g_2$ has the property that $g_2\circ f=g_2\circ g$, but it doesn't satisfy the rest of the properties. What distinguishes $q$ from $g_2$ (thus, $Q$ from $Q_2$) is that for $q:Y\to Q$ there is not necessarily a unique $u':Q_2\to Q$ such, that the proper arrow equality will hold. That is because $g_2$ is not a coequalizer. So, $g_2$ doesn't satisfy the universal property of the coequalizer.
Nice answer but just to make things more clear. In terms of explicit sets and functions. $g_2: Y \to Q_2$ will still lead to the quotient set of Y as the element of $Q_2$? Since this is also done by $Y \to Q$ then does $u$ just map equivalence classes into equivalence classes? – user128828 Feb 14 '14 at 18:00
I think so. But, for example, for some $Q_i$, the arrows $u_i:Q\to Q_i$ might not be surjective. For those, there won't be any $u_i':Q_i\to Q$ with the wanted properties. So actually $Q$ is the set with all equivalence classes of $Y$. – frabala Feb 14 '14 at 18:06