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There are (at least) two definitions of groups in category theory:

  • As a group object (in a catgory $C$ with finite products, e.g. $C$ = Sets). This is a tuple $(G,m,inv,e)$ with the following data satisfying the usual axioms

    • $G$ an object of $C$
    • $m \colon G \times G \to G$ the multiplication morphism
    • $inv \colon G \to G$ the inversion morphism and
    • $e \colon \,\bullet\, \to G$ the identity element morphism, where $\,\bullet\,$ is an terminal object in $C$.
  • As a category with only one object where every morphism is an isomorphism.

My questions are:

  1. As far as I know group objects in Set are equivalent to the second definition. How can I formalize this equivalence in terms of category theory ?

  2. Is there an analoge of the second definition that corresponds to group objects in other categories, e.g. topological groups?

  3. For group objects in Set, it is known that the inverses and identity are uniquely determined if they exist. So they do not have to be included into the data of a group. Does this hold for group objects in every category C?

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    $\begingroup$ (1) You can define the notion of a category object and what it means to have only one object and every morphism is an isomorphism. (2) Yes. (3) Yes. $\endgroup$ – Zhen Lin Jun 27 '15 at 10:50
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  1. If $C$ is a category with finite limits, you can make sense of a category object internal to $C$. You can also make sense of a groupoid object internal to $C$; just add a morphism which plays the role of the inversion. A group object internal to $C$ is then a groupoid object $G$ such that $\mathrm{Ob}(G)=1$, the terminal object.

  2. Isn't this covered by 1.?

  3. Yes. This follows for example directly from the functorial description of groups objects: A group structure on an object $X$ corresponds to a group structure on each hom-set $\hom(T,X)$ for test objects $T$, in such a way that $\hom(T,X) \to \hom(S,X)$ is a group homomorphism for all test morphisms $S \to T$. This can also be used to derive identities such as $(a \cdot b)^{-1} = b^{-1} \cdot a^{-1}$ in the context of group objects.

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  • $\begingroup$ Thank you, 2. is in fact covered by your answer to 1. $\endgroup$ – legacytron Jun 27 '15 at 13:37

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