Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $Q_2$?

share|improve this question
    
What distinguishes them is that there is a canonical arrow $u : Q \to Q_2$ but none in the opposite direction. –  Adeel Feb 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 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 at 17:41
add comment

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
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 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 at 18:06
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.