isn't (some version of) redshift that K raises telescopic complexity (by 1)?

that's not really how < works... :)

what is "telescopic complexity"?

there's a definition at the top of these notes from an MSRI talk by Clark Barwick: msri.org/workshops/689/schedules/18234

but, say, a spectrum E has telescopic complexity $\leq n$ at p if for for any finite complex V of type $k \geq n$, the map $\pi_* V\wedge E \to \pi_*T\wedge E$ is iso, where T is the telescope of the $v_k$ self-map on V.

oops, apparently only iso for $*$ large

does anyone know some reasonable terminology for distinguishing the conditions "Hom(c,-) is faithful" and "Hom(c, -) is conservative"?

@QiaochuYuan first one is "c is a separator/generator", second one is basically "c is a strong generator"

hmm. guess it's time to learn about strong epimorphisms

there are some mild hypotheses, if I recall – you probably need pullbacks and equalisers

here is a question I'm having an astonishingly hard time answering: can every abelian group be written as a colimit colim_{j \in J} F(j) where each F(j) is isomorphic to Z? note that on the one hand it does not suffice to observe that every abelian group has a presentation, but on the other hand 1) every finitely generated abelian group can be written this way, and 2) so can every localization of Z

@QiaochuYuan How would you obtain, say, $\mathbb{R}$ as such a colimit?

can I be the stupid one to ask why a presentation doesn't work? coequalizers seem like colimits to me...

@Dylan I think that the problem is that you have represented every module as an iterated colimit (one step to get the free modules and then another step to get the coequalizer) and it is not immediately obvious how to smash the two colimits together

ah, I see- the map giving the relations isn't a `colimit' of maps, it's a limit of maps

@Qiaochu Z^2 is a dense generator in Ab, so every object A of Ab can be obtained as a colimit of Z^2's in the following way: X :=Hom(-,A) is a presheaf on the one-object category {Z^2} containing Z^2. So the colimit of the inclusion {Z^2} \to Ab, weighted by the presheaf X, is A. Or, if you prefer, there's a natural "evaluation" functor from the category of elements of X to Ab, and its (ordinary) colimit is A.

I'm not sure if this can be done with Z though.

Although Z is dense as an Ab-enriched subcategory of Ab.

So if by "colimit" you allow us to mean "Ab-enriched weighted colimit"

then the answer is yes.

(When I said "I'm not sure if this can be done with Z", well, since Z is not dense in Ab in the Set-enriched sense, certainly the same procedure doesn't work. I'm not sure if there's another procedure though.)

the version with Ab-enriched colimits is trivial, though – just take the tensor!

oh yeah!

@Espen: the abelian groups you can get this way are closed under coproduct and R is a coproduct of copies of Q, so it suffices to explain how you get Q this way. Q is the colimit of the diagram ... -> Z -> Z -> Z where arrow n (from right to left) is multiplication by, say, n

the presentations you can use are presentations which say that one generator is a multiple of another. by the structure theorem this is enough to get all finitely generated free modules (I expect this won't generalize to a more general R and R-modules), and by multiplying by a multiplicative subset of Z you can get localizations of Z in a way generalizing the above; I think a similar trick gets you, for example, Q/Z

an example of an abelian group I don't know how to present this way is the p-adics

@Qiaochu The diagram you just described has colimit Z not Q.

mike shulman claims that this is possible in a note linked to here:

Nevermind I got what you mean

Sorry I'm slow

There is a short exact sequence 0 -> A -> Z_p -> B -> 0, where A is the group Z_{(p)} and B is a rational vector space.

Let's suppose that Z_p were such a colimit. Then Z_p could be written as having a presentation as follows:

It would have a set of generators e_i (indexed by objects in the diagram), and it would have a set of relations all of the form n e_i = e_j (indexed by morphisms in the diagram).

Then I would be able to define a self-map of Z_p as follows:

If e_i is in A = Z_{(p)}, I define f(e_i) = e_i

If e_i is not in A, I define f(e_i) = 0

Then we have to check that this respects the equivalence relation, so we need n f(e_i) = f(e_j)

If n is not zero, then either both e_i and e_j are in A or they are not, because both A and B are torsion-free

If n is equal to zero, then e_j = 0 (hence in A) and f(e_j) = e_j = 0

Therefore this gives a well-defined such map f.

However, any abelian group homomorphism Z_p -> Z_p which is the identity on Z_{(p)} must be the identity (because it must be the identity mod p^n for all n)

Sorry, what is Z_{(p)}?

The p-local integers {a/b | a,b integers, p does not divide b}

Ah, of course. Thanks.

@Tyler: cool, looks good to me! do you want to post it in response to the question I linked to above?

Looking at the above, it seems like it's saying that if M is such a colimit, any surjection from M onto a torsion-free group must split

@QiaochuYuan unfortunately I've got to run and teach for now. if you want to post it, feel free

@TylerLawson: I don't understand why your map f restricts to the identity on Z_{(p)}. What if none of the generators lie in A?

@TimCampion Ah, good point. So I'd need to start with a set of generators, and pick a generator e that mapped to something nontrivial in Z/p; then e would generate the copy of Z_{(p)} that I should use.

In particular, the splitting statement above is much too strong.

Generate how? We don't have a ring structure to work with...

The ring Z_{(p)} acts on Z_p and so I mean the set {x e | x in Z_{(p)}}

So the goal is to have f restrict to the identity on that copy of Z_{(p)} sitting in Z_p?

Yes, that's right.

So... it still might be the case that none of the other generators lie in this copy of Z_{(p)}...

Yes, true.