Working with separation axioms of topological spaces is pretty wild...
They're a bunch of conditions that don't really seem to have any uniform behavior as the index increases, at least from a category theoretic point of view.
@JonathanBeardsley Do you mean something different than the non-colimit definition in ABGHR, namely Definition 2.3: $Mf + \Sigma^\infty_+ P \wedge_{\Sigma^\infty_+ GL_1R} R$, where P is the fibre of $f$?
@JonathanBeardsley $BG$ is a groupoid and the only invertible morphisms in $\Delta^{op}$ are identities, so I don't think so..
@skd Presumably you know this, but Senger also asks this question as Question 1.5 in arxiv.org/pdf/1710.09822.pdf Without saying much else he speculates that the answer is yes. He'd perhaps be a good person to ask in person.
@SeanTilson What is the classical comparison theorem? When C is an abelian category the proposition seems to be more or less saying that any two maps from a bounded below free complex to a free resolution are homotopic. Is this what you were thinking about?
Right, that does seem to be what would happen in an abelian category. The classical comparison theorem, say 2.2.6 of Weibel, constructs an actual map of complexes (up to homotopy) from a map on the two objects the complexes abut to (I have omitted the "obvious" hypotheses).
@OmarAntolín-Camarena To get the "Bousfield-Kan formula" for a colimit you want to restrict from C to (Delta/C)^op along the first (or last?) vertex map, which is cofinal, and then do a left Kan extension along the projection (Delta/C)^op -> Delta^op. Here Delta/C -> Delta is, among other things, the right fibration for C, regarded as a Segal space.
@OmarAntolín-Camarena so i have a kind of naive solution to this, which might be wrong. we have a map $f:BG\to BGL_1(R)$, and since $BG=colim(Bar(\ast,G,\ast))$, it induces maps from every level of that diagram, giving me the bar construction over $BGL_1(R)$. Now use the fact that Thom spectrum construction preserves coproducts, to get that $colim(f)=colim(Bar(R,R[G],R))$. Is that nonsense?
If I have a reasonable space X, with fundamental group G, then I get a map f : X -> BG. Now suppose I have a local system L on X. Am I correct in thinking that R^1f_*L = 0, because the fibre of f is simply-connected?
It feels like I need a proper base change theorem, and f is not proper. So if my idea is wrong: is there a way to compute R^1f_*L ?
@JonathanBeardsley I'm confused. Aren't the n-simplices of Bar(pt, G, pt) given by a product G^n? How are you using that the Thom spectrum construction preserves coproducts?
@jmc So, the map f:X→BG is only defined up to homotopy (in fact BG itself is only defined up to homotopy), while I have the feeling that Rf_*L depends on more than just the homotopy class of f. In fact it should depend exactly on the shape of X relative to BG and that's a geometric notion, not a homotopic one
May and Sigursdsson give a description of the Thom spectrum as a kind of paramterized two-sided bar construction, and I wanted to see if this manifested in the ABGHR description
