Timeline for Could groups be used instead of sets as a foundation of mathematics?
Current License: CC BY-SA 4.0
34 events
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| Jun 28, 2023 at 14:52 | history | edited | John Baez | CC BY-SA 4.0 |
"Martin Brandenbourgh" should be "Martin Brandenburg" in two places; "were" should be "where" in two places.
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| Nov 15, 2022 at 6:40 | vote | accept | Oscar Cunningham | ||
| Nov 15, 2022 at 0:36 | comment | added | Simon Henry | @OscarCunningham The solution proposed by Martin Brandenburg in his answer already solve the problem. But I have edited to include his argument and put everything together. | |
| Nov 15, 2022 at 0:35 | comment | added | Simon Henry | @MartinBrandenburg : thanks I have edited ! I had missed that the the group unit also satisfies the condition that $\Delta(x) =x_L * x_R$. | |
| Nov 15, 2022 at 0:34 | history | edited | Simon Henry | CC BY-SA 4.0 |
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| Nov 14, 2022 at 6:34 | comment | added | Oscar Cunningham | I've removed the green tick until we get an answer with all the right pieces in it. | |
| Nov 14, 2022 at 2:10 | comment | added | Martin Brandenburg | @SimonHenry I would suggest to edit your answer (see my answer). | |
| Feb 14, 2020 at 7:59 | vote | accept | Oscar Cunningham | ||
| Nov 14, 2022 at 6:32 | |||||
| Feb 14, 2020 at 1:11 | history | edited | David Roberts♦ | CC BY-SA 4.0 |
Gave full reference and MR link to the Kan paper
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| Feb 13, 2020 at 21:52 | history | edited | Simon Henry | CC BY-SA 4.0 |
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| Feb 13, 2020 at 21:33 | comment | added | user44143 | HJRW, yes, I am grateful to all of you for pointing out all the subtleties. It’d be nice to see a direct definition of this category, but I may not be the best person to write out all the details. | |
| Feb 13, 2020 at 21:04 | comment | added | HJRW | @MattF., your characterisation of free groups is not correct. Every group that surjects $\mathbb{Z}$ is "almost free" in your sense. (Group theorists call this property "indicable".) There are many "almost free" groups that aren't free and yet don't split as a direct product; a simple example is $\mathbb{Z}*\mathbb{Z}/2\mathbb{Z}$. (And your claim about when maps of free groups take generators to generators is incorrect too, I'm afraid.) Bases of free groups are quite subtle! The situation isn't at all analogous to what happens in free abelian groups. | |
| Feb 13, 2020 at 21:02 | comment | added | HJRW | @SimonHenry, thanks for the explanation. This is very interesting to me; I've thought a lot about bases for free groups, but never seen them re-interpreted in this way. | |
| Feb 13, 2020 at 20:33 | comment | added | Simon Henry | @MattF. : what would be missing is what you say is a purely categorical definition of what it means for a map of free group to sends "generators to generators." (I'm not sure whether your characterization of free group is correct, but I'm assuming it is). And with this sort of definition it would be hard because your characterization of free groups do not fix the set of generators (while the one in terms of cogroups structures does). | |
| Feb 13, 2020 at 19:44 | comment | added | user44143 | Now that I understand this...would the following work as a simplification? 1) A group $G$ is free iff it is almost free (as we figured out in my answer) and not a direct product. 2) A map $f$ of free groups from $G$ to $H$ takes generators to generators iff $H$ is the coproduct of $f(G)$ with some other subgroup. 3) Now consider the subcategory of groups whose objects are free groups and whose morphisms are the maps that take generators to generators. | |
| Feb 13, 2020 at 19:12 | comment | added | user44143 | Thanks! And now I see that the conditions in the ...'s are described in theorem 2.2 in the cogroup link at the nLab, | |
| Feb 13, 2020 at 18:23 | comment | added | Simon Henry | "For all $T$ and $S$ two objects, $\mu_T:T \rightarrow T*T$ a map satisfying ..., $\mu_S:S \rightarrow S * S$ satisfying..., and$f: S \rightarrow T$ satisfying ... ". Of course $T * T$ is not part of the language either, so if I want to be fully rigorous I should add "given an object $T * T$ endowed with two maps $T \rightrightarrows T *T $ satisfying the universal property of a coproduct..." And the same for $S* S$ and $T* T *T$ and $S*S*S$ (because they are needed in the formulation of associativity of the cogroups operations") | |
| Feb 13, 2020 at 18:16 | comment | added | user44143 | It's the quantification over co-group objects and morphisms that I don't understand. How does one write out "Given $T$ a co-group object and $S \rightarrow T$ a co-group morphism"? | |
| Feb 13, 2020 at 17:51 | comment | added | Simon Henry | @MattF. : done in the edit. | |
| Feb 13, 2020 at 17:51 | history | edited | Simon Henry | CC BY-SA 4.0 |
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| Feb 13, 2020 at 17:22 | comment | added | Simon Henry | @HJRW : Cogroup morphisms are not defined as the map the preserves a distinguished subsets, they are defined as the map that are compatible to the co-group structure, i.e. the $f:G \rightarrow H$ such that $\mu_G \circ f = (f * f) \circ \mu_H$ (where $\mu_G$ and $\mu_H$ are the comultiplication maps). It is then a theorem that these are the same as the map preserving the generators. That is what I mean by "reconstructed from the category of groups": we only use concept elementarily formulated in the language of category theory. Otherwise that would not solve the question. | |
| Feb 13, 2020 at 17:12 | comment | added | HJRW | But perhaps that's not the spirit in which the question is meant. I'm not a logician, and neither am I a category-theorist. | |
| Feb 13, 2020 at 17:11 | comment | added | HJRW | @SimonHenry: are you really "reconstructing the category of set from the category of groups"? Or are you reconstructing the category of sets from the category of cogroups inside the category of groups? I ask because the latter seems to me to be much more closely related to sets than the full category of groups. I don't know if this makes sense, but it seems to me that using cogroup morphisms (which preserve a distinguished subset) is really sneaking sets in through the back door. | |
| Feb 13, 2020 at 17:08 | comment | added | Simon Henry | Correction, its not completely an isolated coincidence: it is always the case that given an aglebraic theory T, the free T-models are co-T-models in the category of T-models. But the fact that this induces an equivalence of categories between co-T-models in T-models and sets is not generally true as the example of abelian groups shows. | |
| Feb 13, 2020 at 17:06 | comment | added | Simon Henry | ... the remark by Martin Brandenburg about this specific way of characterizing free groups was very helpful, but this is very specific to the category of group, and I don't see how to generalize this... Also the fact that the trick was to look at "cogroup object in the category of group" is probably an isolated coincidence (for example, every abelian group is automatically an abelian cogroup in the category of abelian group). | |
| Feb 13, 2020 at 17:02 | comment | added | Simon Henry | @HJRW : I don't really know, but that's a great question. As pointed out by Todd Trimble in the comment to the original theory, it is very common that given such a left adjoint, the adjunction is comonadic (essentially, as soon as the left adjoint is non-degenerated enough). But while comonadicity was a very natural way to say that "one can reconstruct the category of set from the category of groups", it was not completely clear that this was enough to solve the problem, as you also need to be able to construct the comonad using only the language of categories. This was the part where... | |
| Feb 13, 2020 at 16:56 | comment | added | HJRW | @SimonHenry. Many thanks. Would this work for any category in which the forgetful functor to set has a left adjoint? | |
| Feb 13, 2020 at 14:37 | comment | added | Simon Henry | When I say "the category of cogroups" I mean the category of object endowed with a cogroup structure, and cogroup morphisms between them (exactly as when you talk about the category of group you mean groups and groups morphisms). So, (as the "theorem" says) free groups equiped with a basis. | |
| Feb 13, 2020 at 8:44 | comment | added | HJRW | Just to make sure I understand correctly, are your objects here co-groups (ie free groups) or groups equipped with a co-group structure (ie free groups with a fixed basis)? | |
| Feb 13, 2020 at 3:09 | comment | added | David Roberts♦ | This is really nice! | |
| Feb 12, 2020 at 23:25 | history | edited | user44143 | CC BY-SA 4.0 |
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| Feb 12, 2020 at 22:21 | history | edited | Simon Henry | CC BY-SA 4.0 |
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| Feb 12, 2020 at 22:14 | history | edited | Simon Henry | CC BY-SA 4.0 |
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| Feb 12, 2020 at 22:03 | history | answered | Simon Henry | CC BY-SA 4.0 |