@SaalHardali Here's another class of examples: If $C$ is finitely (co)complete and there is an equivalence $C = C^{op}$, then $Ind(C)$ is self-dual in $Pres^L$. For $Ind(C) \otimes D = Fun^R(Ind(C)^{op},D) = Fun^{lex}(C^{op},D) = Fun^{rex}(C,D) = Pres^L(Ind(C),D)$, naturally in $D$.

Scratch that.

Actually, I think something like Yonatan Harpaz's list of examples must be all there is. At least, I think any category with a dual must be a colocalization of a presheaf category. For $Ind(C)$ is a localization of $P(C)$, so by duality we expect $Ind(C)^\vee$ to be a colocalization of $P(C^{op})$. This would follow if we knew that the tensor of presentable categories is 2-functorial in arbitrary accessible functors, so preserves adjunctions.

let $M$ be a commutative monoid. is its classifying space $BM$ necessarily a 1-type?

i don't see why this should be true, and yet it feels like it is

@AaronMazel-Gee you're assuming the discrete topology on M, right? (Otherwise you have BU(1) for a counterexample)

@ArunDebray yes sorry, i meant a discrete commutative monoid

if nothing else i feel like this should follow from some sort of "calculus of left fractions" type of argument (the key point being that everything commutes, so the relevant condition is trivially satisfied)

but that's not so conceptually satisfying

in case it's helpful, i'd be willing to assume the "right cancellability" that appears in the nlab page: ncatlab.org/nlab/show/calculus+of+fractions

which in turn implies that dwyer--kan's stuff in "calculating simplicial localizations" can also be applied

@SaalHardali I guess I am using the definition of formal schemes in Rezk's supplementary notes. So I think spf(R), where the ideal I is understood, should be pro-represented by the system {R/I^n} in the adic rings. The category of adic A-algebras is defined by Rezk to be the category whose objects are A-algebras B with augmentations B--> A with nilpotent kernel

So I guess if k is a field, I think of Spf(k) as being "pro-represented" by {k/0^n} since the only ideal is 0.

Wait, there are formal schemes without points? that's weird. What's a canonical example?

@AaronMazel-Gee Yes, when M is commutative the canonical map BM --> BG is a homotopy equivalence, where G is the Grothendieck group of M. You can prove this easily using Quillen's theorem A, and as you surmised the key point is that M has a calculus of left fractions (in particular, you don't need any cancellation property). So indeed, any monoid M with such a calculus of left (or right) fractions satisfies that BM --> BG is an equivalence, with G = algebraic group completion of M.

@AaronMazel-Gee This seems to be true and is proven in math.leidenuniv.nl/scripties/LenzMaster.pdf as 4.3.1

People that study monoids and semigroups call having a left (or right) calculus of fractions "satisfying the left (or right) Ore condition", I believe.

@OmarAntolín-Camarena aha, i like your argument better than mine. in particular, it implies finality! that's a very nice result to know: if $G = M^{gp}$ and $G$ acts on $X$, then $X_{hM} \xrightarrow{\sim} X_{hG}$. and the same argument implies initiality, so that also $X^{hM} \xleftarrow{\sim} X^{hG}$.

@PiotrPstrągowski thanks for the reference. it's always nice to see intuition corroborated through other means, but i can't say that i personally consider what's there as "conceptual" (but surely that is my own ignorance)

@OmarAntolín-Camarena hmm, although actually there's something subtle going on. it's by definition that $BM \to (BM)^{gpd}$ is a localization, hence both initial and final. i guess the nontrivial input here is the equivalence between ($\infty$-)monoids and pointed ($\infty$-)categories with a single equivalence class of object.

so weirdly, we're proving that for $G = M^{gp}$ the ordinary group-completion $BM \to BG$ is a weak equivalence by proving that it's final, and then using the above to deduce that actually $G$ is the $\infty$-group completion, from which it follows that $BM \to BG$ is a localization.

another random question. suppose i have a group G (abelian if you like). for what sorts of spaces $X$ is the mapping space $hom(BG,X)$ just equivalent to $X$? (basically equivalently (i believe), for what sorts of pointed spaces $Y$ is the mapping space $hom_*(BG,Y)$ contractible?) for instance, interesting cases might include $G = \mathbb{Z}/p$, or more generally $G = \mathbb{Z}/n$, or in another direction $G = \mathbb{Q}$.

this feels a little funky because we're mapping into $X$, whereas e.g. i believe that rational homotopy theory only applies in the case of $G = \mathbb{Q}$ for mapping out of $X$

@AaronMazel-Gee That's the Sullivan conjecture, isn't it?

@YonatanHarpaz That's interesting, considering I arrived to this question by thinking about cosheaf and sheaf categories and how they sit inside presentable categories. Is it true for any topos continuous presheaves on it gives back the same topos? I was thinking that in general one can define sheaves on a topos with values in $\mathcal{C}$ as continuous functors $\mathfrak{X}^op \to \mathcal{C}$ and cosheaves as cocontinuous functors $\mathfrak{X} \to \mathcal{C}$....

... Is this the right perspective?

@CWcx I really don't know terribly a lot about formal schemes and I wish I had phrased my previous comment differently, let me try again:

My point is that you should either have a notion of a formal scheme with no points or you should live with the fact that closed points (or generally closed subsets) cannot be removed from formal schemes. Unlike in the category of schemes.

For example take the formal disk $D := Spf \mathbb{C}[[x]]$ this has one closed point and no other points. Sometimes we might like to consider also the punctured formal disk $D^*$ as some object with a morphism $D^* \to D$. But if $D^*$ has a point then it has to go to the only closed point in $D$ and so $D^*$ can't possibly be punctured.

I guess it's a matter of terminology, since the punctured formal disk is naturally a pro-ind scheme and not an ind-scheme. So whether you include those or not is up to you. My point was intended to be about this phenomenon that you can't remove points and you don't have nice generization morphisms like in the classical case.

@SaalHardali, I think this perspective is correct. In particular, suppose that your oo-topos is sheaves on some site C. Then indeed LFun(Sh(C),Spaces) = coSh(C), and so if Sh(C) is dualizable then its dual should be coSh(C). However, to have duality you would also need to have that Sh(C) = Fun(coSh(C),Spaces), and I'm not sure how to show that (or even if it's true in this generality).

The only case where I know how to show this is the very special "projective" case. This is the case where the localization PSh(C) -> Sh(C) is not just left exact, but actually preserves all limits. This happens, for example, if every object has an initial covering sieve (for example, if you take a compact Hausdorff space X and consider a sieve on U to be covering if it contains all V -> U in which the closure of V belongs to U).

I call it projective because in this case Sh(C) is a retract of the free presentable oo-category PSh(C) in Pr^L. In fact, in this case both Sh(C) and coSh(C) will be projective (hence both will be oo-topoi) and one can show that Sh(C) = LFun(coSh(C),Spaces).

@DenisNardin haha yes, right of course -- at least for finite groups. but what about something like $G = \mathbb{Q}^\times$?

@MarcHoyois I think what I forgot is that you should also "homotopify" by modding out concordance (i.e. make R conctractible). I don't have a reference, but I believe I heard this in connection with the Stolz-Teichner stuff on field theories, so it might be discussed in one of their papers...

Hello

I was wondering if this question classefies to be posted in mathoverflow ?

@RuneHaugseng hi

so, i know lots of examples of spaces where it's nice to compute their homology / cohomology by using mayer-vietoris and expressing them as a union. does anyone know examples of spaces where it's nice to compute their homology / cohomology by using mayer-vietoris and expressing them as an intersection?

maybe a smooth complete intersection in CP^n? but ideally i'd want an example which isn't already given to you as an intersection

@QiaochuYuan there's the homological proof of the Jordan Curve theorem, which I really like

first, you show that for any embedding [a,b] -> S^2 with image K, the reduced homology of S^2 \ K vanishes

proof: divide K into the images K_1 of [a,(a+b)/2] and K_2 of [(a+b)/2,b]

then you get a Mayer-Vietoris in reduced homology .... -> H_*(S^2 \ K) -> H_*(S^2 \ K_1) + H_*(S^2 \ K_2) -> 0 -> ...

(because the union is S^2 \ {p})

if H_*(S^2 \ K) is nonzero, then so is some H_*(S^2 \ K_i). but then you can keep repeating that until K_i is arbitrarily small, and you can get a linear contraction of any cycle you had to start with

second, you assume that you have some embedding S^1 -> S^2, let U be the image of the upper hemisphere, L the image of the lower, with overlap {p,q}

then you get a Mayer-Vietoris in reduced homology ... -> H_*(S^2 \ S^1) -> 0 -> H_*(S^2 \ {p,q}) -> ...

which shows reduced H_0(S^2 \ S^1) is Z, thus you have two path components

and the second part of the argument uses MV to calculate the homology of an intersection

my recollection is that the same technique can be used to prove some variant of Alexander duality for embeddings of finite CW-complexes, where the "unions" on the CW-complex side turn into "intersections" on their complements

(maybe that needs to be simplicial complexes)