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Epimorphism is defined as following: $f \in \operatorname{Hom}_C(A,B)$ is epimorphism if $\forall Z. \forall h', h'' \in \operatorname{Hom}_C(B, Z)$ the following holds: $h' f = h'' f \; \Rightarrow \; h' = h''$.

I can't think of examples where epimorphism would not have a right inverse.

Also, if I understand correctly, epimorphism is not surjective in the categories where we can't talk about surjection (where objects does not have internal structure?).

Thanks in advance.

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Take the category of a partially ordered set. Every arrow is an epimorphism, but no non-identity arrow has a right inverse. – Arturo Magidin Nov 11 '11 at 14:51
Try playing with (Hausdorff) topological spaces and continuous maps, for example. The map $x \mapsto e^{ix}$ from $\mathbb{R} \to S^1$ is surely epi but has no continuous right inverse. A morphism is epi if and only if it has dense range. – t.b. Nov 11 '11 at 14:52
In the category of rings, epimorphisms do not, in general, have right inverses. For example, $f:\mathbb Z \rightarrow \mathbb Z/<p>$ has no right inverse. – Thomas Andrews Nov 11 '11 at 14:53
In some categories, "morphisms" are not represented as functions of sets, so we can't say a morphism is "surjective," in general, only that it is epimorphic. – Thomas Andrews Nov 11 '11 at 14:55
In the category of rings an epimorphism does not even have to be surjective. Take the natural embedding of $\mathbb{Z}$ in $\mathbb{Q}$. Since $\mathbb{Q}=Quot(\mathbb{Z})$ every morphism with domain $\mathbb{Q}$ is already defined by its values on $\mathbb{Z}$. Therefore the embedding is a non-surjective epimorphism. – Sebastian Schoennenbeck Nov 11 '11 at 15:09
up vote 18 down vote accepted

Take the category of a partially ordered set; every arrow is an epimorphism, but no non-identity arrow has a right inverse.

For a concrete category (objects are sets and morphisms are functions between the underlying sets), take the category of Hausdorff topological spaces; an epimorphism is a continuous map with dense image. Consider $\mathbb{Q}\hookrightarrow \mathbb{R}$. This is an epimorphism, but there is no retract (no right inverse). Or the map $[0,2\pi)\to S^1$ given by $t\mapsto (\cos t,\sin t)$. If it had a right inverse in the category, the inverse would be a bijection, hence we would have homeomorphisms, but $[0,2\pi)$ and $S^1$ are not homeomorphic.

For yet another, take $\mathbb{Z}\hookrightarrow\mathbb{Q}$ in the category of rings with unity. This is an epimorphism, but does not have a right inverse.

Even in concrete categories where all epis are surjective, you need not have right inverses. In the category of groups, an epimorphism $G\to K$ has a right inverse if and only if $G$ is a semidirect product $G\cong N\rtimes K$. So take $\mathbb{Z}/4\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$ for a surjective morphism (hence an epi) with no right inverse in the category.

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+1 for elaborating on differtn types of counterexamples. – Hagen von Eitzen Sep 12 '12 at 6:09

Let $C$ be the category containing one object $\mathbb Z$, and morphisms being functions $f_n(z)=nz$ for $n\in\mathbb Z^+$. Then $f_n\circ f_m = f_{nm}$, and we can see that:

$$f_n\circ f_k = f_n\circ f_l \implies f_k=f_l$$

And similarly:

$$f_k\circ f_n = f_l\circ f_n \implies f_k=f_l$$

So every $f_n$ is an epimorphism and a monomorphism, but only $f_1$ has a left or right inverse. Note that $f_n$ is never surjective, if $n>1$, even though it is an epimorphism.

If you defined $C'$ in the same way, but with the object being $\mathbb Q$ and, for $n\in\mathbb Z^+$, $f_n(q)=nq$ is defined as a function on $\mathbb Q$, then this new category is (in some sense) isomorphic to $C$, and all functions are surjective and injective.

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Let $ \textbf{Ring}_{\text{Assoc}} $ be the category of associative (not necessarily unital) rings. We take the morphisms in this category to be maps between associative rings that preserve ring addition and ring multiplication. Note that it is irrelevant to speak of unit-preserving maps in this category as rings may not be unital.

Now, let $ i: \mathbb{Z} \rightarrow \mathbb{Q} $ be the inclusion map. We claim that it is an epimorphism. Let $ f,g: \mathbb{Q} \rightarrow R $ be two morphisms from $ \mathbb{Q} $ to another ring $ R $ such that $ f \circ i = g \circ i $. Hence, $ f(n) = g(n) $ for all $ n \in \mathbb{Z} $. Let $ \dfrac{a}{b} \in \mathbb{Q} $. Then \begin{align} f \left( \frac{a}{b} \right) &= f \left( \frac{1}{b} \cdot a \right) \\ &= f \left( \frac{1}{b} \right) *_{R} f(a) \\ &= f \left( \frac{1}{b} \right) *_{R} g(a) \\ &= f \left( \frac{1}{b} \right) *_{R} g \left( b \cdot \frac{a}{b} \right) \\ &= f \left( \frac{1}{b} \right) *_{R} \left[ g(b) *_{R} g \left( \frac{a}{b} \right) \right] \\ &= f \left( \frac{1}{b} \right) *_{R} \left[ f(b) *_{R} g \left( \frac{a}{b} \right) \right] \\ &= \left[ f \left( \frac{1}{b} \right) *_{R} f(b) \right] *_{R} g \left( \frac{a}{b} \right) \\ &= f \left( \frac{1}{b} \cdot b \right) *_{R} g \left( \frac{a}{b} \right) \\ &= f(1) *_{R} g \left( \frac{a}{b} \right) \\ &= g(1) *_{R} g \left( \frac{a}{b} \right) \\ &= g \left( 1 \cdot \frac{a}{b} \right) \\ &= g \left( \frac{a}{b} \right). \end{align} Therefore, $ f = g $, which implies that $ i: \mathbb{Z} \rightarrow \mathbb{Q} $ is indeed an epimorphism. This result strengthens the one mentioned by Arturo above.

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