i'm most skeptical of my identification $Sp^{hG} \simeq Mack(Burn(GFin^{free}))$, and now i'm thinking that actually we only have $Sp^{hG} \simeq Fun^\oplus((GFin^{free})^{op},Sp)$ -- is that the entire issue, the additive transfers?

That identification is true (it's been proven by Saul)

But I do not see how to identify either of your functors as RKE

In particular it is not clear to me why (-)^G is the RKE along (-)^G

well, it's right-adjoint to a functor called $j^*$ -- isn't that pretty suggestive?

@DenisNardin which identification?

The identification of Sp^{hG} with Mackey functors on the Burnside category on free G-sets

I do not see why (-)^G is right adjoint to j^*

Maybe I am missing something obvious. What I see is that geometric fixed point is left adjoint to j^*

Also, adjunctions at the level of finite G-sets do not in general induce adjunctions at the level of Mackey functors

applying Fun(-,Sp) to an adjunction gives another adjunction

Yeah, but you need an adjunction at the level of Burnside categories

ah, is there no adjunction on Burnside categories?

Not always, and I believe not in this case

OHHHHH okay

fantastic, thank you

by the way, what is the reference for saul's equivalence $Sp^{hG} \simeq Mack(Burn(GFin^{free}))$?

Appendix A in this paper: arxiv.org/abs/1507.01976

great

by the way, are there conditions under which an adjunction of categories-with-finite-coproducts induces an adjunction on Burnside categories?

In fact you can see that Map_{A(*)}(I,J^G)!=Map_{A(G)}(I,J)

sorry, by != you mean \notequals, right? nothing to do with LKan

it took me a while to parse that

Yes sorry

(and by naturality squares I mean the square that you need to check the commutativity of to see if the (co)unit is a natural transformation)

And of course, you also need to check that your functors preserve cartesian squares (but that's not what fails in this case)

hmm, let me just try unwinding this. you're saying that for a map $S \to T$ in $Fin$, we need $S \to (S \times G/e)^G \to (T \times G/e)^G$ and $S \to T \to (T \times G/e)^G$ to define a pullback square?

oops, wrong functor for left adjoint

You need (S→S^G)→(T→T^G) to be a pullback square (and it is) and (X^G→X)→(Y^G→Y) (and this is not)

yeah so the unit is actually a natural iso, so no problems there

oh, good call on writing commutative squares as arrows of arrows :o)

you need the counit square to be a pullback, really, not a pushout? (i'm not claiming it is, just confirming)

I am pretty sure it is pullback

I am just checking that the adjunction identities hold in the Burnside category

that feels weird, because if we take opposites we should get the same condition (Burn is insensitive to opposites, right)

A(C) is self dual but A(C) is very different from A(C^op)

oh haha sure, a span in C is very different from a span in C^{op}

my mistake

@DenisNardin and what did you mean by this line? which functors, just the adjoints between categories-with-finite-coproducts?

If you have a functor F:C→D you need that it preserves pullbacks if you want a functor F:A(C)→A(D)

otherwise it does not play well with composition

right okay, just making sure

great, thanks again for the help! this genuine-equivariant stuff is really tricky, at least for someone who hasn't thought about it much

It keeps being tricky after you thought about it for a couple of years...

haha well i suppose those statements don't actually contradict each other...

I didn't really read the above discussion but, in case someone is asking, the right adjoint to the inclusion of geometric spectra is, I believe, the 'cogeometric fixed points' given by F(\tilde{EP}, -)^G ? It's not used very often for some reason

also omg... I just checked three different references and found three different conventions for whether i or j should be for open or closed embeddings, and it's making me insane. left and right is confusing enough without also having to keep track of this

I'm this close to calling open embeddings "Open" and closed embeddings "Closed" but I think that would upset everyone... including myself.

The open embedding is j. Please do not use any other convention...

But be careful that the role of the open and closed embedding swap between constructible sheaves and quasi-coherent sheaves (that might be the reason for differing conventions)

@Drew A Bousfield localization $f:X\to Y$ is characterized by (i) $Y$ is $B$-local, i.e., there are no maps of $B$-acyclics to $Y$, and (ii) the fiber of $f$ is $B$-acylic. Just apply this directly to $X\to B\wedge X$ when $B$ is smashing.

BTW, for my original question: assume $B$ is a unital $A$-module which is smashing (whence $B$ is an $A$-algebra) and dualizable. Let $F=$ fiber of $A\to B$, which will also be dualizable since $A$ and $B$ are. Then the dual of $F\to A$ is a map $A\to F^*$, which will be also be smashing, whence $F^*$ is an $A$-algebra. ....

Then $F^*\otimes_A B=0$, because this is the same as the function $A$-module $[F,B]$, which is contractible since any $F\otimes X\to B$ factors through $B\otimes F\otimes X=0\otimes X$. Thus $A\to B\times F^*$ is smashing, and is also conservative by a similar argument, so is an equivalence by @DenisNardin's argument.

Sorry to ask such a simple question, but does anyone have a reference for the spectra X_i, defined in page 2 of this pdfs.semanticscholar.org/1251/…

They are defined as the Thom spaces of the maps f_i:\Omega S^i\to BO, which are given as the loops of generators of the homotopy groups of B^2O. I ask as I am unable to verify the properties listed in section 2.