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I would like to solve the Problem 2.12 in Szekeres, A Course in Modern Mathematical Physics:

Show that the class of groups as objects with homomorphisms between groups as morphisms forms a category – the category of groups (see Section 1.7). What are the monomorphisms, epimorphisms and isomorphisms of this category?

In the first part, I consider category $\mathbf{Grp}$ as the class of all groups and the group homomorphism as category morphism, and I show the needed properties.

The confusing part is to find the monomorphisms, epimorphisms and isomorphisms of this category.

In https://en.wikipedia.org/wiki/Group_homomorphism we have the associations of monomorphism with injective, epimorphism with surjective, and isomorphism with bijective. But in https://en.wikipedia.org/wiki/Homomorphism#Category_theory in paragraph in about Category theory I read:

However, the definitions in category theory are somewhat different.

I interpret the sentence to means: Category monomorphism, epimorphism, and isomorphism are different than the usual homomorphism definitions (for rings, groups, aso) Is this true for groups?

Concretely I have found no way to derive, for instance, that monomorphisms from the group category are injective homomorphism. I mean: starting from:

The monomorphisms of $\mathbf{Grp}$ are the homomorphisms $\phi:G\to H$ such that

$\forall X\in\mathbf{Grp},\forall \alpha,\alpha':X\to G$, we have: $\phi\circ\alpha=\phi\circ\alpha'\implies \alpha=\alpha'$

And get to $\phi(x)=\phi(y)\implies x=y$.

Another observation. In Szekeres the Problem is stated after very few definitions (and no theorem!). Thus I cannot use things like functors as in Is every monomorphism an injection?

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2 Answers 2

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It is obvious that injective (respectively, surjective) group homomorphisms are monomorphisms (respectively, epimorphisms).

Monomorphisms in the category of groups are injective maps. Indeed, suppose $\phi\colon G\to H$ is a monomorphism; consider $\alpha\colon\ker\phi\to G$, the canonical injection, and $\beta\colon\ker\phi\to G$, $\beta(x)=1$. Then $\phi\circ\alpha=\phi\circ\beta$: what does $\alpha=\beta$ entail?

Epimorphisms in the category of groups are surjective, but this is a bit more difficult to show (one needs to define an action on the set of cosets of $H$ by the image of $\phi$).

The standard example of a nonsurjective epimorphism in a category is the embedding $\mathbb{Z}\to\mathbb{Q}$ in the category of rings, which is both a monomorphism (obvious) and an epimorphism (try it).

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  • $\begingroup$ A proof of the fact that a group epi is surjective is to be found in Carl Linderholm, "A Group Epimorphism is Surjective", American Mathematical Monthly 77, pp. 176-177. From the article (paraphrased): Of the 6 assertions (mono iff injective, epi iff surjective, iso iff bijective) the only one that is not "very easy" is epi $\implies$ surjective. $\endgroup$ Jul 14, 2016 at 18:43
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There is a more categorical proof of the fact that the underlying function of an epi group-morphism is surjective. The first step is to note that the monomorphisms in the category of groups are all equalizers, in fact every monomorphism is the equalizer of its cokernel pair. The proof is constructively valid and pretty straight forward. Also it is the natural thing to try, since if a monomorphism has a cokernel pair and is an equalizer, then it is automatically the equalizer of its cokernel pair.

Take an epimorphism $e:G \to H$. Consider the factorisation $G \to e(G) \to H$ of $e$ through its image. Clearly the underlying set map of $G \to e(G)$ is surjective, so it is enough to show that the inclusion $e(G) \to H$ is an isomorphism. The map is monic, because its underlying set map is. Because of the above it is regular monic. It is also an epimorphism, because its composite with $G \to e(G)$ is epi. A standard argument shows that every morphism which is both epi and regular monic is an isomorphism, so we are done.

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