An epimorphism is defined as follows:

$f \in \operatorname{Hom}_C(A,B)$ is an epimorphism if $\forall Z, \forall h', h'' \in \operatorname{Hom}_C(B, Z)$ then $h' f = h'' f \; \Rightarrow \; h' = h''$.

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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    $\begingroup$ Take the category of a partially ordered set. Every arrow is an epimorphism, but no non-identity arrow has a right inverse. $\endgroup$ Nov 11, 2011 at 14:51
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    $\begingroup$ 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. $\endgroup$
    – t.b.
    Nov 11, 2011 at 14:52
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    $\begingroup$ 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. $\endgroup$ Nov 11, 2011 at 14:53
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    $\begingroup$ 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. $\endgroup$ Nov 11, 2011 at 14:55
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    $\begingroup$ 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. $\endgroup$ Nov 11, 2011 at 15:09

5 Answers 5


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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    $\begingroup$ +1 for elaborating on differtn types of counterexamples. $\endgroup$ Sep 12, 2012 at 6:09
  • $\begingroup$ So the part "with unity" can be dropped in the third paragraph by your other answer. $\endgroup$
    – user557
    Jul 26, 2019 at 2:46
  • $\begingroup$ @user634426: Those are two different assertions: it is an epimorphism in the category if rings with unity; it is also an epimorphism in the category of rings that do not necessary have unities. Both are correct. $\endgroup$ Jul 26, 2019 at 3:17
  • $\begingroup$ @ArturoMagidin I thought that the former implies the latter, but apparently it doesn't. $\endgroup$
    – user557
    Jul 26, 2019 at 3:19
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    $\begingroup$ @user634426: Well, both are true, so the implication holds; but you cannot deduce the latter from the former “categorically” (if $\mathbf{C}$ is a subcategory of $\mathbf{D}$ it does not follow that if $f$ is an epi in $\mathbf{C}$ then it is also an epi in $\mathbf{D}$, even if $\mathbf{C}$ is a full subcategory. The implication does hold in the opposite direction; here we only need an example, not the most general possible example. $\endgroup$ Jul 26, 2019 at 3:23

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.

  • $\begingroup$ This is in a way an application of the universal property of the field of fractions. Any canonical inclusion of an integral domain that isn't a field into its field of fractions should give analogous examples. $\endgroup$ Jan 29, 2022 at 14:08

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.


I should have known this for years, but I actually learned it only today:

Let $K$ be a field of characteristic $0$. In the category of finite-dimensional (!) Lie algebras, the inclusion map of the standard Borel subalgebra

$$\{\pmatrix{a&b\\0&-a}: a,b \in K \}$$

into $$\mathfrak{sl}_2(K) = \{\pmatrix{a&b\\c&-a}: a,b,c \in K \}$$

is an epimorphism (and obviously not surjective). Actually, so is the inclusion of any Borel (and hence any parabolic) subalgebra of a split semisimple Lie algebra.

Note that in the category of all Lie algebras over a given field, indeed epimorphisms are surjective, but one really needs infinite-dimensional Lie algebras for this.


In the category of monoids, the inclusion map $\iota:\mathbb N\to\mathbb Z$ is an epimorphism, but it is obviously not surjective. (Here, I consider $0$ to be a natural number.)

To prove that it is an epimorphism, suppose that $M$ is a monoid, and $\alpha,\beta:\mathbb Z\to M$ are monoid homomorphisms such that $\alpha\circ \iota=\beta\circ\iota$. Since $M$ is not necessarily commutative, I will use multiplicative notation for it.

By hypothesis, $\alpha|_{\mathbb N}=\beta|_{\mathbb N}$. Let $n\in\mathbb N$. Note that $\alpha(n)$ is invertible; indeed, \begin{align} \alpha(n)\alpha(-n)=\alpha(-n)\alpha(n)=\alpha(0)=1 \, , \end{align} whence $\alpha(-n)=\alpha(n)^{-1}$. Similarly, $\beta(-n)=\beta(n)^{-1}$. Thus, $$ \alpha(-n)=\alpha(n)^{-1}=\beta(n)^{-1}=\beta(-n) \, , $$ and so $\alpha$ and $\beta$ agree on the nonpositive integers, too. Hence, $\alpha=\beta$.

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    $\begingroup$ The codomain of $\alpha$ and $\beta$ could be any arbitrary monoid, not just $\mathbb{N}$. Also, you are begging the question when you say that a monoid homomorphism with a domain of $\mathbb{Z}$ is determined by its values on $\mathbb{N}$. $\endgroup$ Jul 27, 2023 at 20:41
  • $\begingroup$ You still need to show why monoid homomorphisms with a domain of $\mathbb{Z}$ are determined by their values on $\mathbb{N}$. $\endgroup$ Jul 30, 2023 at 15:46
  • $\begingroup$ @GeoffreyTrang: Okay, I updated my answer. What do you think now? $\endgroup$
    – Joe
    Jul 30, 2023 at 16:00
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    $\begingroup$ That will work. Thanks, Joe. $\endgroup$ Jul 30, 2023 at 16:18

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