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It seems like I do understand what it means - to be "unique up to isomorphism". I'll try to formulate brief yet rigorous explanation (rather than just definition) in terms of equivalence relations.

So, let's take something really simple. How about toy-set $S = \{ \frac{1}{2}, \frac{2}{3}, \frac{3}{4} \}$ containing only those 3 elements and good-old well-know equivalence relation $(=)$?

(1) There are infinitely many bijections (also known as 'isomporhisms') from $f : S \mapsto S'$, for example: $S' = \{ \frac{2}{4}, \frac{4}{6}, \frac{6}{8} \}$, which, gathered all together, form yet another set $Iso(S)$.

(2) moreover: $\forall f \in Iso(S), \forall (s \in S, f(s)): s = f(s)$, i.e. defined equivalence relation holds.

In other words, none $S, S'$ are exactly identical, but all of them preserve somewhat particular equivalence relation. And that's why those are said to be unique up to a bijection (or an 'isomorphism'): no narrower uniqueness exists.

Did I miss something important?

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    $\begingroup$ $Iso(S)$ would be a proper class. I don't know what you're trying to do with (2). Where does $s'$ come from? What equivalence relation was "defined"? I think you're making this more complicated than it is. $\endgroup$ Jan 30 '18 at 19:49
  • $\begingroup$ @DerekElkins, typo: $s = f(s)$, where $=$ denotes somewhat equivalence relation between pair $(s, f(s))$. For simplicity sake, I proceeded on the ordinary 'equals' relation. $\endgroup$
    – Zazeil
    Jan 30 '18 at 19:52
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    $\begingroup$ What is the question? $\endgroup$
    – user332239
    Jan 30 '18 at 19:54
  • $\begingroup$ @user332239 the question is: do I understand things correctly? Is that a valid way of thinking about "uniqueness up to isomorphism"? $\endgroup$
    – Zazeil
    Jan 30 '18 at 19:55
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    $\begingroup$ @MorganRodgers as far as I understand, my toy example deals with bijections preserving defined equivalence relation. And that's one aspect of the entire question - does it make sense to think in such a way? May be my way of thinking is not quite correct, since there is better point of view to see the "big picture"? $\endgroup$
    – Zazeil
    Jan 30 '18 at 19:58
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It doesn't really make sense to talk about "unique up to isomorphism" without some context. The way it virtually always shows up is you have some particular property, $P$, and you want to say any $X$ such that $P(X)$ holds is unique up to isomorphism. This simply means that for any other $Y$ for which $P(Y)$ holds, $X\cong Y$. If we wanted to say that given any $X$ such that $P(X)$ holds was simply unique, we'd have a formula like $\forall X,Y.(P(X)\land P(Y))\implies X=Y$. For unique up to isomorphism, this just becomes $\forall X,Y.(P(X)\land P(Y))\implies X\cong Y$.

In many cases in category theory, we actually have that things are "unique up to unique isomorphism", which simply means that there is exactly one isomorphism $X\cong Y$ when $P(X)$ and $P(Y)$ both hold.

As a final caveat, usually what the property $P$ is being applied to is not a single object but rather a collection of objects and arrows, e.g. the categorical product is talking about an object and two arrows (the projections). However, it is common to talk just about the object part and leave the arrows implicit. Related to this, we are usually considering isomorphisms in categories other than $\mathbf{Set}$, e.g. for the categorical product we'd be considering a category of cones, say.

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  • $\begingroup$ Looks like the only difference between my understanding and your explanation is that you allow any property $P(x)$ which should be "preserved" by appropriate bijections. However, my way of thinking was slightly less generic: equivalence relations is kind of $P(x, y)$ indeed (i.e. both before-bijection and after-bijection values are equivalent), but $P(x)$ shouldn't be limited to equivalence relations. Hopefully, I got your idea correctly. $\endgroup$
    – Zazeil
    Jan 30 '18 at 20:05
  • $\begingroup$ An isomorphism "preserving" a property $P$ would be something like: if $P(X)$ holds and $X\cong Y$, then $P(Y)$ holds. This is a very different situation. I think you are just confused about the "equivalence relation" stuff. It's neither relevant to this concept, nor, as far as I can make out, able to fit as an example. For any $s\in S$, $f(s)$ can be any set whatsoever, so your relation would need to be a proper class. $\endgroup$ Jan 31 '18 at 5:26
  • $\begingroup$ @DerekElkins seems like having $X \cong Y$ is sufficient to build up an equivalence relation. Let say there is a relation $R$ such that $P(X) \land P(Y) \Rightarrow (X, Y) \in R$. It seems to meet all the equivalence requirements: reflexive, symmetric, transitive. $\endgroup$
    – 52heartz
    Jan 31 '18 at 8:01
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Derek Elkin makes the key points. But just for fun, let's play with your example a moment. You asked us to consider the set:

$S = \{ \frac{1}{2}, \frac{2}{3}, \frac{3}{4} \}$

Suppose I'm cheerfully unco-operative and say "Stop right there! What set is that? The elements are supposed to be rational numbers -- but what are they? Are they ordered pairs of ordered pairs of naturals? That's a standard story! In which case how do you define ordered pairs -- you can do that set theoretically in all kinds of ways ...."

You will, in this context, quite rightly respond "Oh come off it! Don't be awkward!! That's irrelevant to what I wanted to say about this set and the other ones I mentioned. Pick whichever story about the rationals you like, or treat them as unanalysed primitives, and start from there ...!!!"

"Ahah," I reply, "You are of course right. The differences between the different implementations just aren't relevant to what you wanted to do with the example. OK, why not? Because -- on the different implementations -- the whole structure you get for the "rationals" of the different flavours is isomorphic."

"Does that mean that the rationals in the different stories are the same?"

"No, we can't say that, strictly speaking. But since we in this case don't care about the differences, they are as-good-as-the-same, as far as the structural properties that are going to matter are concerned. Or, as they say, looking across the implementations, you do get the same things up to isomorphism."

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    $\begingroup$ Your post is great and funny, but making next step closer to the essence, how can one ever figure out that something is 'as-good-as-the-same' without proper equivalence relation? I got no way to compare things bypassing such a step, do I? $\endgroup$
    – Zazeil
    Jan 30 '18 at 20:17
  • $\begingroup$ You need a proper isomorphism between relevant structures. Not any old equivalence relation between them. (E.g. cardinal equivalence by itself isn't, normally enough for a relevant isomorphism -- though that equivalence by itself will be entailed by the isomorphism.) $\endgroup$ Jan 30 '18 at 20:27
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If I understand your toy example, I think you are saying two sets $S$ and $S^{\prime}$ are isomorphic iff there is a bijection between $S$ and $S^{\prime}$ (any bijection at all).

In this case, we could say a set $S$ of cardinality 3 is unique up to isomorphism, because any other set $S^{\prime}$ of cardinality 3 is isomorphic to $S$.

Note that we need this caveat "of cardinality 3": It would not make sense to simply say that $S$ is unique up to isomorphism, because there are sets which are not isomorphic to $S$ (any set $S^{\prime}$ with a different cardinality, for example).

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