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I know that in the category of groups, the epimorphisms are precisely the surjective homomorphisms. What about the category of monoids? One can easily see, that surjective homomorphisms are epic (even effectively epic).

Edit: Denoting the left adjoint $\mathbf{Mon}\longrightarrow\mathbf{Grp}$ of the forgetful functor $U$ by $G$, the unit of the adjunction $M\longrightarrow UGM$ is always epic, though not always surjective, as the embedding $\Bbb{N}\longrightarrow\Bbb{Z}$ shows.

The surjective homomorphisms should coincide with the effective epimorphisms.

But: Is there a useful characterization of epimorphisms?

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    $\begingroup$ There's a counterexample right in the Wikipedia page... Did you really search? $\endgroup$ – Najib Idrissi Mar 6 '15 at 12:16
  • $\begingroup$ @NajibIdrissi I added the result to my question. $\endgroup$ – user114885 Mar 6 '15 at 12:22
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This is an interesting question. I don't know any "useful" and "practical" characterization, but one can say something about epis in $\mathsf{Mon}$:

Since the "universal enveloping group" $G: \mathsf{Mon} \to \mathsf{Grp}$ is left adjoint, it preserves epimorphisms, and epimorphisms in $\mathsf{Grp}$ are known to be surjective. Thus, if $f : M \to N$ is an epimorphism of monoids, then $G(f) : G(M) \to G(N)$ is surjective. Equivalently, for every $n \in N$ the induced element $\underline{n} \in G(N)$ has some preimage in $G(M)$, say $\underline{m_1} \cdot \underline{m_2}^{-1} \cdot \dotsc \cdot \underline{m_k}$. This means that $\underline{n} = \underline{f(m_1)} \cdot \underline{f(m_2)}^{-1} \cdot \dotsc \cdot \underline{f(m_k)}$ in $G(N)$. I don't know how to make this more explicit; in general it seems to be hard to tell if two elements of $G(N)$ are equal.

Here is another idea: If $M \to N$ is an epimorphism of monoids and $M \twoheadrightarrow M' \hookrightarrow N$ is its usual (regular epi, mono)-factorization, then $M' \to N$ is an epimorphism, and the converse is also true. Thus, in order to characterize epimorphisms, we may assume that $M \hookrightarrow N$ is a submonoid. There is a general result stating that a morphism $f : X \to Y$ in a category with pushouts is an epimorphism if and only if the canonical "codiagonal" map $Y \sqcup_{X} Y \to Y$ is an isomorphism. So we would be done if we can describe $N \sqcup_M N$ for a submonoid $M \hookrightarrow N$.

The element structure of the coproduct $N \xrightarrow{\iota_1} N \sqcup N \xleftarrow{\iota_2} N$ looks as expected: Elements can be written as finite "alternating" products $ \dotsc \cdot \iota_1(u) \cdot \iota_2(v) \cdot \iota_1(w) \cdot \dotsc$ with unique elements $\dotsc,u,v,w,\dotsc$ from $N \setminus \{1\}$. The pushout $N \sqcup_M N$ is the quotient of $N \sqcup N$ by the smallest congruence relation $\sim$ such that $\iota_1(m) \sim \iota_2(m)$ for $m \in M$. So how to make $\sim$ explicit? I don't think that this will be easy. In the case of groups, there is a well-known element structure of pushouts using coset representatives, but this is not available for monoids.

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