Uniqueness up to isomorphism?

Question

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?

Answer 1

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.

Answer 2

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."

Answer 3

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).