Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Assume $f,g:X \to Y$ are arrows in $\mathsf{Sets}$. Then the coequalizer is given by $c:Y \rightarrow Y/R$ where $R \subseteq Y\times Y$ is the smalles equivalence relation on $Y$ s.t. $\forall x \in X: (f(x),g(x)) \in R$. Given any $h:Y \rightarrow Z$ there is a unique $\overline h: Y/R \to Z$ s.t. $\overline h \circ c = h$. I know that $\overline h ([y]) = h(y)$.

My question: How can I prove that $[y]=[y'] \Rightarrow h(y) = h(y')$ ?

share|cite|improve this question
What is $c$? You have not defined it. And if it is a quotient map then $h$ would be also forced to be equal on objects of the same equivalence class. I misunderstand something – porton Oct 26 '13 at 20:18
@porton: He forgot to mention that $hf$ must be the same as $hg$, see my answer below. – Stefan Hamcke Oct 27 '13 at 22:35
up vote 2 down vote accepted

Given any $h:Y\to Z$ such that $hf=hg$ ...

Now, "have the same image under $h$" determines an equivalence relation $H$, and since $(f(x),g(x))\in H$ it follows that $R\subseteq H$.

share|cite|improve this answer
For clarity: Make $H \subseteq Y^2$ equivalence relation s.t. $(y,y') \in H \iff h(y) = h(y')$. Then since $hf = hg$ we have $\forall x \in X: (f(x),g(x)) \in H$. With the minimality of $R$ we know $R \subseteq H$. Then if $[y]=[y']$ we have $(y,y') \in R$ and thus $(y,y') \in H$ s.t $h(y) = h(y')$. Right ? – Epsilon Oct 10 '13 at 15:58
@André: Yes, that's exactly how it works. – Stefan Hamcke Oct 10 '13 at 16:01
Thanks. It seemed strange to me that those minimal equivalence relations exist. But now I recognize that the intersection of eq. relations is again an equivalence relation and $\mathcal P(Y^2)$ is always one. – Epsilon Oct 10 '13 at 16:06
@André: Not $\mathcal P(Y^2)$ but $Y^2$. – Stefan Hamcke Oct 10 '13 at 16:15
Oops yes, sure. – Epsilon Oct 10 '13 at 19:06

Your Answer


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.