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I am a novice in the field of category theory, and one of the things I struggle to wrap my head around is the notion of universal properties. Precisely, I struggle to understand why universal properties all seem to be stated in terms of the existence of a unique morphism between objects, instead of just at least one morphism.

Now, I understand that the very idea of a universal property is to define an object up to isomorphism via a certain property. In a sense, this property becomes the definition of the object. But more specifically, universal properties define objects up to unique isomorphism. What I don't understand is why we want unique isomorphisms between these objects, instead of at least one isomorphism. What would be lost by not having a unique isomorphism between objects that satisfy a property?

Note that I understand how requiring a unique isomorphism means that objects that satisfy the same universal property are isomorphic. But is it necessary?

I've read this question whose top rated answer explains the terminology, but doesn't really explain why uniqueness up to unique isomorphism is useful or interesting or desirable. There is also this question that is very similar to mine, and whose answer tries to justify why the unicity of the isomorphism matters, but I'm not really convinced by the explanation. Again, wouldn't simply an isomorphism uniquely characterise the object in question?

Edit 1:

It has been pointed out to me in the comments that universal properties state that there exists a unique morphism that makes a certain diagram commute, and that it is the uniqueness of this morphism that is important. Not so much the fact that it gives unique isomorphisms between satisfying objects.

Thinking about it more, I realised that if the morphism given by a universal property wasn't unique, then the objects that satisfy that property wouldn't necessarily be isomorphic to one another. Is this the reason why the uniqueness of that morphism is important?

Thomas.M
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  • The example in your second link is not a good one, in that, as pointed out in one of the comments there, the uniqueness in the universal property of fields of fractions is not actually necessary. But that is a special case. Your question is a bit mixed up about the role of uniqueness: what is asserted to be unique is the morphism to or from the universal object that makes a certain diagram commute and it is the uniqueness of that morphism that is important. The fact that this gives unique isomorphisms between different candidates for the universal objects is useful but not central. – Rob Arthan Nov 14 '23 at 21:18
  • @RobArthan Thank you for your comment. Yes, I could have worded my question more clearly. I will try to rephrase it better. I was indeed asking about the uniqueness of the morphism. You say that it is the important part of universal properties, which I can easily believe, but I'm not sure I understand why. Then again, without it, it seems that objects that satisfy the same universal property are not necessarily isomorphic, so maybe this is why the uniqueness of the morphism is important? – Thomas.M Nov 14 '23 at 21:32
  • Slightly disagreeing with @RobArthan, in my own experience, the uniqueness-up-to-unique-isomorphism of a thing specified by a universal property is very important: two different constructions (e.g., to prove existence) invariably yield the same thing. "Things are what they have to be." :) True, this disallows "automorphisms" of universal objects (in a certain sense), but, upon closer examination, that's usually fine. :) The thing that finally convinced me about this uniqueness feature was W. Rudin's "definition" of the (appropriate) topology on distributions... seemed... [cont'd] – paul garrett Nov 14 '23 at 22:02
  • ... [cont'd] unmotivated and needlessly messy, until I realized that he was describing a construction of a colimit, etc. :) Then proved (as a sequence of unexplained lemmas) the properties of a colimit. :) – paul garrett Nov 14 '23 at 22:03
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    One thing that points to the importance of the uniqueness of the morphisms: In applications, we will very often want to use that unique morphism in further constructions. And generally, mathematicians are much more comfortable with the notion of taking "the unique thing which satisfies some property" than with the notion of "choose some arbitrary thing out of the collection of multiple things which satisfy some property, and fix that choice for all time". – Daniel Schepler Nov 14 '23 at 22:17
  • @paulgarrett: I really meant central to the universal property as a concept: I agree that uniqueness of the isomorphisms between different candidates is a very important consequence of the universal property. – Rob Arthan Nov 14 '23 at 22:34
  • @RobArthan, I suspected so, but didn't want anyone to over-interpret your remark. :) – paul garrett Nov 14 '23 at 22:36

1 Answers1

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You already noted that the idea of universal properties is to characterize an object by how it relates to other objects. For definiteness, let's look at the universal property of the coproduct of topological spaces. Given two spaces $X$ and $Y$, the idea is to characterize the coproduct $X\sqcup Y$ by stating that, for any space $Z$, a map $X\sqcup Y\to Z$ should be the same data as two maps $X\to Z$ and $Y\to Z$. It is this idea of ''the same data'' that forces us to require uniqueness of the morphism in the universal property. To see this, suppose for a moment that we remove uniqueness from the universal property of the product. Then (if $X$ and $Y$ are nonempty) any space of the form $X\sqcup Y\sqcup W$ also satisfies this modified universal property: two morphisms $X\to Z$ and $Y\to Z$ give us possibly many different ways to define a map $T\colon = X\sqcup Y\sqcup W\to Z$ extending our given maps. This means that specifying a map out of $T$ requires an unspecified amount of extra data than just two maps out of $X$ and $Y$. Indeed, we may vary our choice of $W$ and will then still satisfy the modified universal property, so we don't even know what extra data we need to specify a map out of $T$. So without uniqueness in the universal property, we don't actually know how an object satisfying it relates to other objects, which defeates the point of a universal property. Without uniqueness, we only have a very general idea that mapping in or out of it requires more data than something else, but we just can't work with this amount of vagueness.

As for the question why we would like unique isomorphisms between objects, let me answer a slightly different question: why do we want preferred/canonical isomorphisms between objects, rather than just any isomorphism? The reason is that even though isomorphisms allow us to see two objects as similar, it can be dangerous to actually identify them without having a canonical isomorphism. I will give three examples.

  1. We do not actually identify in general a finite-dimensional vector space $V$ with its dual $V^*$ even though they are abstractly isomorphic, because there are many incompatible choices for such an isomorphism and no preferred one. If you choose for any vector space $V$ an isomorphism $\varphi_V\colon V\to V^*$, then in some sense you are picking a specific basis of $V$ to work in, instead of any other basis. So if you decide to identify $V$ and $V^*$ using $\varphi_V$, this means that you are in a sense not working in a theory of linear algebra which studies intrinsic properties of vector spaces, but in a theory that studies also properties coming from a possibly weird (and incompatible between different spaces) choice of bases.
  2. Consider three spaces $X$, $Y$ and $Z$, and the two products $(X\times Y)\times Z$ and $X\times(Y\times Z)$. We often just write $X\times Y\times Z$ for either space and we will not run into problems doing this. The reason is that there is a preferred and natural isomorphism $\alpha_{X,Y,Z}\colon (X\times Y)\times Z\to X\times(Y\times Z)$ between both objects. If you are mean, you could give for a specific $X$, $Y$ and $Z$ a stupid isomorphism between these objects, but once we are going to look at products of four spaces, say $X\times Y\times Z\times W$, your stupid isomorphisms will not produce a single isomorphism between $(X\times Y)\times (Z\times W)$ and $((X\times Y)\times Z)\times W)$, but will produce multiple different isomorphisms between these. This is because there are different ways to use associativity to move from one way of bracketing to another way of bracketing. In other words, using stupid isomorphisms to obtain associativity in the product of three spaces, you don't know anymore how to identify different bracketings in products of four spaces, and this is a problem. The problem is completely resolved if you just stick to your preferred isomorphisms $\alpha_{X,Y,Z}$ that the universal property of the product gave you.
  3. Automorphisms of an object are its symmetries. If you could just use any isomorphism to identify two objects, you are sort of saying that we can pretend that any symmetry of an object is the identity symmetry. This is not true (we can often deduce information about an object by studying interesting symmetries, which would not be possible if all symmetries were essentially just the identity symmetry).

The unique isomorphisms between objects satisfying a universal property are needed because they give us canonical isomorphisms between these objects, and thus allow us to identify all objects satisfying that universal property. This allows us to pretend there is just one empty space, and just one one-point space, and just one product of two spaces, and allows us to deduce (coherent) symmetry and associativity of the product of spaces, etc.

Daniël Apol
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    Great answer +1. I will point out that what you say about products becomes nuanced in categories like the homotopy category, where the objects are sets and the product objects are product sets, but the morphisms aren't just special kinds of set-theoretic functions. This leads to the fascinating world of Stasheff polytopes. – Rob Arthan Nov 14 '23 at 22:48
  • This is a very clear and enlightening answer, and it clears my conceptual doubts on the matter. Thank you for the effort you put into writing this thorough answer! – Thomas.M Nov 24 '23 at 01:29