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In his answer to Unique up to unique isomorphism, Qiaochu Yuan explains quite well what is meant by something being "unique up to a unique isomorphism", but I'm a bit perplexed by the significance of the uniqueness of this isomorphism. What makes "unique up to unique isomorphism" more useful than merely "unique up to isomorphism"?

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Something can be significant without being "useful," i.e. it can inform our perspective and understanding and intuition and appreciation without being used as a lemma in proofs. For example, Cayley's theorem. (Although its generalization, the Yoneda lemma, is very useful.) The idea of category theory is to think in terms of arrows, not elements, and so it makes sense to talk about unique arrows as much as it does to talk about unique objects. –  anon Dec 18 '13 at 17:42
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1 Answer 1

"Unique up to unique isomorphism" is significant because not only is the object itself uniquely identified, but the individual elements are as well.

For example, $\mathbb{Z}$ as an additive group is not unique up to unique isomorphism, because we cannot distinguish 1 from -1. This means that any place a group isomorphic to $\mathbb{Z}$ arises, we will always have a choice of generator. In the absence of additional information, there will be no natural way to decide which element is 1 and which is -1.

However, $\mathbb{Z}$ as a ring is unique up to unique isomorphism. With multiplication, we can distinguish 1 from -1. Whenever a ring isomorphic to $\mathbb{Z}$ arises, not only do we identify the ring itself, but also individual elements we can label 0,1,2,3,... and -1,-2,-3,...

Edit: Here is perhaps a better example. All vector spaces over a fixed field $F$ of a fixed dimension $n$ are isomorphic. However, this isomorphism is highly non-unique, relying on a choice of basis. This tells us that we should not usually think of any such vector space $V$ as simply being elements of $F^n$, because there is no natural choice of which object of $V$ should be (1,0,...,0), etc.

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How does it help us to be able to do that for the ring? Or how does having two distinct generators for the group hurt us? –  dfeuer Dec 18 '13 at 17:11
    
@dfeuer In the case of algebraic topology, a local orientation of a topological manifold $M$ of dimension $n$ at a neighbourhood of a point $p$ is determined by a choice of generator of the homology group $H_n(M, M\setminus \{p\}) \approx \Bbb Z$. So the two possible choices of generators correspond to the two possible local orientations. –  Arthur Dec 18 '13 at 17:17
    
@dfeuer I added another another example, which is perhaps better than the $\mathbb{Z}$ example. –  Ted Dec 18 '13 at 17:35
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