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I’ve realised that in my article recently posted to the cafe, I make some assumptions about the existence of some pullbacks right before I show said pullbacks are isomorphic to a known object $P$ (making that known object actually the pullback). At first glance it seems wrong, but then if I think about it in terms of generalised elements (writing them as ordinary elements), I think I’ve actually shown that
$Hom(-,A)\times_{Hom(-,C)}Hom(-,B) \simeq Hom(-,P)$hence the LHS is representable, and so the pullback exists and is equal to $P$. Is this a valid argument?
I’m pretty sure that is valid by Yoneda’s lemma and/or the definition of a limit.
I know the second part is valid, but what about the first version? Hmm, reading over my post I wasn’t too clear. I’ll say it like this:
Can I pretend that the pullback exists, write down an explicit isomorphism between P and the pullback using the various maps available, and then claim I’ve just secretly argued existence by the ’real’ argument, that of showing the LHS in #1 is representable/use Yoneda etc?
This is purely diagrammatic reasoning, and I guess it’s kind of a metaquestion with a philosophical bent. If I assume that some object exists, then show this other thing I have is isomorphic to it, then does the original thing exist? I’m thinking in a kosher fashion here, where really I can substitute my ’other thing’ for the one I originally only assumed to exist.
Yes, that’s exactly what the limit is. Take a look at the page on the nLab: limit.
I think that in absolute generality, the answer can’t be more than “sometimes.” I’m not thinking of any off the top of my head, but I’m pretty sure I’ve run into situations where I was only able to prove that a given object had some universal property by assuming that there was some object with that universal property and constructing an isomorphism. However, it’s true that often such arguments can be rephrased by showing directly that the object which is known to exist does have the desired universal property. For instance, if your argument could be reinterpreted as taking place in the presheaf category, where all small limits do exist, then what you’ve shown is that the limit presheaf is in fact representable, and since the Yoneda embedding reflects limits (essentially by definition of “limit”), it follows that the given object is in fact the limit.
For instance, if your argument could be reinterpreted as taking place in the presheaf category, where all small limits do exist, then what you’ve shown is that the limit presheaf is in fact representable, and since the Yoneda embedding reflects limits (essentially by definition of “limit”), it follows that the given object is in fact the limit.
This is what I was getting at.
@Harry - I know what a limit is :) My question is more ’would people buy it?’
@Mike - I think that’s what I was getting at. I originally wrote out the proof in question (of something else - the pullback was just a step) assuming existence of limits, so I don’t think it unreasonable to work in the category of presheaves and reflect back.
@David: I realize that you understand what a limit is. I was just directing you to the part of that page that discusses defining limits to be objects representing “formal limits” in the presheaf category.
@Harry, that’s ok. :)
There is a criticism of category theorists that I have heard voiced (in a low voice of course) and that is that at the drop of a hat they throw in assumptions that all lifts exist and they are often not necessary. The representability of a functor as described in #1 is the way that the early Grothendieck school of category theory defined the limit to exist, so I do not really think there is any problem. Anyhow from the iso in #1 you can prove the universal property for P directly, surely!
An example where I would like to prove that something has a universal property but cannot: Assume that a pretopos has inductive and coinductive objects (initial algebras and final coalgebras of polynomial functors); in particular, the natural-numbers object $\mathbf{N}$ (the initial algebra of $X \mapsto X + 1$) and the stream object $Str(A)$ (the final coalgebra of $X \mapsto A \times X$) for each object $A$. If I assume that the exponential object $A^{\mathbf{N}}$ exists, then I can prove that $Str(A)$ is isomorphic to it, but I can’t prove that $Str(A)$ is an exponential object without an existence assumption.
@Tim: Is there any other way to define the existence of the limit?
”Not really” or ,”of course not”, depending on how you are feeling.
But sometimes one needs a specific limit or colimit to exist (just that one itsi-bitsi object to be there) and in reply to a question one gets the reply, why don’t you assume the category has all limits … or similar. The passage to presheaves is the same argument, but it is interesting (and reasonably instructive) to glance back at the categorical reasoning of Grothendieck etc. in the 1960s. In a lot of the early moduli space stuff he clearly does not really worry about whether the space exists as a scheme/algebraic space/stack, or not, but feels that he has to produce one out of the hat to show it ’exists’ so as to satisfy the geometric public!
(A little story, not completely relevant!: We had a visitor from another UK university over to give a seminar, and his speciality was moduli spaces and their existence. He gave a very good talk for a non-specialist audience. I asked him at the end, why was it useful or important to know that moduli spaces of that type ’existed’. He told me the main reason was that then they would be able to calculate the cohomological invariants of the moduli spaces, which held important geometrical information. I was fairly certain that the ’existence’ of the moduli space was not needed in order to calculate its cohomology. (i.e. the cohomology existed even if the space did not!) He agreed with my intuition. The point was that a functor existed that, if representable, gave the moduli space as its representing object, but once you had the moduli space most people would go back towards the functor or related constructs that could be worked on without the space, but, as he explained to me, geometers liked to feel the cohomology was of something. The existence was therefore in many ways a psychological prop to aid mathematicians to push things further. I don’t discard that as a valuable point although it may seem a bit silly. I could give other instances of the same thing happening in lots of other parts of maths.)
Isn’t a moduli stack essentially defined to be the functor one desires it to represent?
Yes. but the incident that I related is at least twenty five years ago, probably 1985 or there abouts. At that time the moduli space was thought of as a space. The notion of scheme was acceptable as a geometric object (but only recently) and the algebraic spaces of Artin were looked at with suspicion. Deligne and Mumford in the Zariski volume (Pub. IHES, 1969) had treated a moduli space both as a generalisation of a ringed space and as an (algebraic) stack (almost shock horror). People still wanted the space there.
Another instance of the same thing was when I defined the Cech homotopy type/ shape to be a pro-space (defined up to pro-homotopy). People more or less said that that could not be correct as you had to have a space there, not a whole system of spaces. (The modern point would be to look at the functor prorepresented by the pro-space). There are still people who think in the old way and have great difficulty in seeing that the various ways are equivalent at the elementary level whilst the newer ways score better when trying to get new and deeper results.
The problem is at the same time logical and psychological! What does it mean for a limit to exist and why do we want it to? If we work with the corresponding presheaves, isn’t that as good? I suspect the answer to the conundrum is ’sometimes yes, sometimes no!’ (a true Norman response, if you know the tradition interpretation of that in France.)
Very interesting.
a true Norman response, if you know the tradition interpretation of that in France
The Internet is letting me down here.
From the French wikipedia page:
« Répondre en Normand » : ne répondre ni oui ni non, comme dans l’adage caricatural « p’têt ben qu’oui, p’têt ben qu’non ».
« Réponse normande » : réponse exprimée en termes ambigus.
Thanks; serves me right for missing the “d”.
You can find a neat example in Asterix!! I leave you to find which!! :-)
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