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

Let $M$ be a monoid and consider the category of $M$-acts. The morphisms of this category are mappings that preserve the action of $M$. Let $f : X \rightarrow Y$ be an epimorphism in this category, i.e. a right-cancellable morphism. I am trying to prove that actually $f$ is a surjective mapping, but i am having difficulty. Thanks for your help.

share|cite|improve this question
Why not try proving a stronger statement? For any category $\mathcal{C}$, a morphism (i.e. natural transformation) in the functor category $[\mathcal{C}, \textbf{Set}]$ is an epimorphism if and only if it is a componentwise surjection. Hint: Yoneda lemma. – Zhen Lin Mar 4 '12 at 18:47
@ZhenLin: What do you mean by "functor category"? – Manos Mar 4 '12 at 21:01
The category of all functors $\mathcal{C} \to \textbf{Set}$. A $M$-set is a special case of a functor, when you take $\mathcal{C}$ to be a one-object category. – Zhen Lin Mar 4 '12 at 23:42
up vote 2 down vote accepted

Let's assume that $f:X\to Y$ is not surjective, and let's show that $f$ is not an epimorphism.

It is straightforward to check that the equivalence relation $\sim$ defined on $Y$ by $$ y\sim y'\iff y,y'\in f(X) $$ is a congruence.

Let $z\in Y/\!\!\sim$ be the equivalence class $f(X)$, and let $c:Y\to Y/\!\!\sim$ be the constant map equal to the equivalence class $f(X)$.

Then we have $p\circ f=c\circ f$ but $p\neq c$.

share|cite|improve this answer
This is great. Thanks Pierre! – Manos Mar 5 '12 at 20:58
Dear @Manos: You're welcome! – Pierre-Yves Gaillard Mar 6 '12 at 1:09

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.