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In my study of Type Theory, the univalence axiom has been introduced as the statement that "Isomorphic structures are equal." I haven't learned how to define any algebraic, combinatorial, or topological structures in type theory yet. BUT I do know that there are times in math where it pays to think of isomorphic structures as distinct representation of the same isomorphism class. For example, even though $2\mathbb{Z} \cong 3\mathbb{Z}$, $\mathbb{Z}/2\mathbb{Z}\ncong\mathbb{Z}/3\mathbb{Z}$.

So, does the univalence axiom imply $\mathbb{Z}/2\mathbb{Z}\cong\mathbb{Z}/3\mathbb{Z}$? My guess is of course not, since this would probably create lots of problems, but then what am I missing in my understanding of the univalence axiom?

Perry Bleiberg
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  • The answer to your question is no, of course. Maybe you can have a look at the first section of my survey preprint (https://drive.google.com/file/d/0B63Vr5tVwiAjdEJhSW03YXBNMVE/view). It should help you. The slogan "isomorphic structures are equal" is a shortcut, hence you may need further explanations. – A. Bordg Aug 21 '17 at 17:54

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In order for $G/H$ to be a well defined object, it's not enough for $H$ to just be a group. It needs to be a (normal) subgroup of $G$. One way to phrase this in type theory, is that $H$ is a group together with an embedding $H \hookrightarrow G$. Then $2 \mathbb{Z}$ and $3 \mathbb{Z}$ are not isomorphic as subgroups of $G$, because an isomorphism would have to preserve all the structure, including the embeddings into $G$.

There is a "forgetful functor" from subgroups of $G$ to groups that just throws away the embedding. The images of $2 \mathbb{Z}$ and $3 \mathbb{Z}$ are then equal, assuming univalence, because they are isomorphic as groups. But that's fine - it just says that the forgetful functor is not injective on objects. Similar things can even happen in set theory.

aws
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  • Thanks so much. – Perry Bleiberg Aug 22 '17 at 00:10
  • What’s the point of declaring “isomorphic structures are equal”, when you have to add more structure in order to make this not run into troubles? Working in set theory, I can always just add the entirety of the background set theoretic structure and just declare that two things are equal when they are isomorphic “as sets”, considering their embedding into the set theoretic universe and all the membership relations in the transitive closure of their images. – mbsq Feb 23 '24 at 08:33
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    It seems very natural to me to require $H$ to be a subgroup of $G$ when defining $G/ H$, so I don't think of it as adding extra structure to avoid trouble, but yes, you technically don't need to do that in set theory. Also in HoTT you could define groups as elements of the cumulative hierarchy together with group structure and do the same thing if you really want to. – aws Feb 23 '24 at 09:42
  • @aws Good point, thanks. – mbsq Feb 23 '24 at 14:59