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An object $G$ in a category $\mathcal{C}$ is called a group object if, given any object $X$ in $\mathcal{C}$, there is a group structure on the morphisms $\operatorname{hom}\left(X,G\right)$ such that $X\mapsto \operatorname{hom}\left(X,G\right)$ is a (contravariant) functor from $\mathcal{C}$ to $\text{Grp}$.

Group objects of the category $\text{Set}$ are groups in the usual sense. Similarly, group objects of the category $\text{FinSet}$ are finite groups.

Assuming that good (non-tautological) descriptions exist,

  1. What are the group objects of the category $\text{FinBij}$, the category whose objects are finite sets, and whose morphisms are bijections?

  2. What are the group objects of the functor category of $\text{FinBij}$, the category whose objects are functors of $\text{FinBij}$, and whose morphisms are natural transformations?

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    $\begingroup$ So basically the consensus is that while your question is perfectly explicit and well-formulated (and the answer is given in Mariano's answer), what you are considering is not what people usually call group objects in $\mathcal{C}$, but rather representable group objects in the category of presheaves on $\mathcal{C^{op}}$. $\endgroup$
    – Captain Lama
    Apr 27, 2016 at 8:44
  • $\begingroup$ Question 2 is not well-defined. You say the objects are the "functors of $\mathsf{FinBij}$": functors have a domain and a codomain, you only stipulate one of them without saying which one. Do you mean endofunctors, that is functors $\mathsf{FinBij} \to \mathsf{FinBij}$ ? $\endgroup$
    – Pece
    Apr 27, 2016 at 15:18

2 Answers 2

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There is no group object: since you only have bijections in your category, for all G there exists an X such that hom(x,G) is empty, so not a group in any way at all.

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    $\begingroup$ By the way, I can also create a definition for which an empty set is a group and says it is not the "classic" definition. $\endgroup$
    – Tsemo Aristide
    Apr 27, 2016 at 1:33
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    $\begingroup$ You surely can, but the OP didn't in his question. If you ask a question and make explicit such a definition, I would probably have to think more. $\endgroup$
    – Mariano Suárez-Álvarez
    Apr 27, 2016 at 1:39
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    $\begingroup$ He didn't precise what the definition of group was for him $\endgroup$
    – Tsemo Aristide
    Apr 27, 2016 at 1:40
  • $\begingroup$ Isn't this answer another reason to not call the OP's definition a "group object"? Bijections of a 1-element set do not constitute a group object (under that definition). $\endgroup$
    – zyx
    Apr 27, 2016 at 1:40
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    $\begingroup$ Why would the set of bijections of a 1-element set not constitute a group?! It is certainly not empty. Not even the set of bijections of a 0-element set is empty! $\endgroup$
    – Mariano Suárez-Álvarez
    Apr 27, 2016 at 1:42
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The question uses a nonstandard definition of group object.

Under the ordinary definition of group object, the answer to the first question is

  1. For the identity morphism $1 \to G$ and the multiplication morphism $G \times G \to G$ (if products exist...) to live in a category whose morphisms are bijections, $G$ must have cardinality $1$.

The definition in the question is equivalent to the usual only when the category has finite products. Although your definition would seem to be more general, in a category with terminal object, the subcategory generated by that object has a unique structure of finite products (all of which are equal to the terminal object) and one would therefore expect that the terminal object fits any definition of "group object". As written in Mariano's answer this is not the case for your definition. Therefore a different phrase than "group object" should be used for this more general definition.

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    $\begingroup$ This has essentially the same problem as Tsemo's answer. $\endgroup$
    – Mariano Suárez-Álvarez
    Apr 27, 2016 at 1:24
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    $\begingroup$ You could argue that his definition is bad or non standard or anything, but he was explicit about what definition he had in mind! $\endgroup$
    – Mariano Suárez-Álvarez
    Apr 27, 2016 at 1:29
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    $\begingroup$ Oh well. I guess you should report it to overloading police or something! ;-) $\endgroup$
    – Mariano Suárez-Álvarez
    Apr 27, 2016 at 1:47
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    $\begingroup$ In fact, I think this definition is not that unusual, althought it is usually only considered in contexts where products do exist. It certainly makes sense in general. And it is a good one: how would you define a "topological space object in a category"? A sensible way is to say that it is an object together T with a factorization of hom(-,T) through the forgetful functor from spaces to sets. In this case, it is not clear what the "internal to the category" version of the definition is, if there is any at all! $\endgroup$
    – Mariano Suárez-Álvarez
    Apr 27, 2016 at 1:52
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    $\begingroup$ @MarianoSuárez-Alvarez It is not the issue but if you are interested: the internal version of a topological space object uses the notion of sketches. Sketches, introduced by Charles Ehresmann and its school, are a kind of generalized Lawvere theories. If I remember correctly, the sketch for topological spaces was introduced by Albert Burroni. $\endgroup$
    – Pece
    Apr 27, 2016 at 15:13

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