15
$\begingroup$

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.

$\endgroup$
  • 8
    $\begingroup$ Take the category of a partially ordered set. Every arrow is an epimorphism, but no non-identity arrow has a right inverse. $\endgroup$ – Arturo Magidin Nov 11 '11 at 14:51
  • 6
    $\begingroup$ 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 '11 at 14:52
  • 2
    $\begingroup$ 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. $\endgroup$ – Thomas Andrews Nov 11 '11 at 14:53
  • 2
    $\begingroup$ 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. $\endgroup$ – Thomas Andrews Nov 11 '11 at 14:55
  • 6
    $\begingroup$ 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. $\endgroup$ – Sebastian Schoennenbeck Nov 11 '11 at 15:09
34
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ +1 for elaborating on differtn types of counterexamples. $\endgroup$ – Hagen von Eitzen Sep 12 '12 at 6:09
  • $\begingroup$ So the part "with unity" can be dropped in the third paragraph by your other answer. $\endgroup$ – user634426 Jul 26 '19 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. $\endgroup$ – Arturo Magidin Jul 26 '19 at 3:17
  • $\begingroup$ @ArturoMagidin I thought that the former implies the latter, but apparently it doesn't. $\endgroup$ – user634426 Jul 26 '19 at 3:19
  • 1
    $\begingroup$ @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. $\endgroup$ – Arturo Magidin Jul 26 '19 at 3:23
7
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

A easier example is to take the category of group.

Consider the inclusion from H={(), (12)} to symmetric group S3. This homomorphism is not surjective, and doesn't have a right inverse. However it is an epimorphism, and its cokernal is trivial as well.

$\endgroup$
  • $\begingroup$ This map is not an epimorphism. In the category of groups, epimorphisms are all surjective. You can see a proof here $\endgroup$ – Arturo Magidin Apr 1 at 15:45
  • $\begingroup$ To see an easy reason why this map is not an epimorphism, consider the two embeddings of $S_3$ into $S_4$, one as the stabilizer of $4$, and one as the stabilizer of $3$. The two maps agree on $H$, but do not agree on all of $S_3$ (in fact, they agree on $H$ and only on $H$), so the embedding cannot be an epimorphism. You do not detect epimorphisms by looking at the normal closure of the image (i.e., the cokernel). $\endgroup$ – Arturo Magidin Apr 1 at 15:55
  • $\begingroup$ Thank you for the reply. I will check again closely if my example is an epimorphism. However, in the textbook I am using it says that your conclusion, that epimorphisms are all surjective, holds only in the category of abelian groups. That's because from epimorphism we can derive the triviality of cokernal, but we can no longer invoke the universal property of cokernals to get to the next step since the normality of image is not guaranteed. On the other hand, the converse, that all surjective group homomorphisms are epimorphisms, is true. $\endgroup$ – Yunhan Apr 3 at 7:42
  • $\begingroup$ I do not know what your book means to say: it is true that you cannot use the cokernel to conclude that epimorphisms are surjective in the category of all groups. However, it is still a fact that epimorphisms are surjective in the category of all groups. I already pointed you to a proof. If your book is insinuating that it is not the case that all epimorphisms are surjective in the category of all groups, then I strongly urge you to throw that book in the garbage right now. $\endgroup$ – Arturo Magidin Apr 3 at 17:09
  • $\begingroup$ Here, again, is the proof that your map is not an epimorphism. Call your embedding $i$. If it were an epimorphism, then whenever $fi = gi$, you would be able to conclude that $f=g$ (epimorphisms are precisely the right cancellable morphisms). Define $f\colon S_3\to S_4$ via the obvious embedding. Define $g\colon S_3\to S_4$ by composing $f$ with conjugation by $(34)$. Then $f$ and $g$ agree precisely in the elements of $S_3$ that fix $3$, which is precisely $H$. That is, $fi=gi$. However, $f\neq g$, because $f(23) = (23)$ but $g(23)=(24)$. So $i$ is not an epimorphism. $\endgroup$ – Arturo Magidin Apr 3 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.