A distinguished triangle of mapping spectra arising from recollement

Question

I suspect the following should be well known, in some circles, under some name. Alas, I could not figure out how to prove it or where to look it up.

Recall that a recollement is a sequence of (triangulated) functors of triangulated categories

$\mathcal{D}' \xrightarrow{i_*} \mathcal{D} \xrightarrow{j^*} \mathcal{D}''$

satisfying certain conditions, see e.g. 1. I just want to recall that $j^*$ is also denoted $j^!$ and has a right adjoint $j_!$, that $i_*$ has a left adjoint $i^*$, that $i_*, j_!$ are fully faithful, that $j^* i_* = 0$ and that for $X \in \mathcal{D}$ there is a (unique, functorial) distinguished triangle $j_! j^! X \to X \to i_* i^* X \xrightarrow{w_X} j_! j^! X[1]$.

Now let us assume that we are in a sufficiently rich situation that there are mapping spectra $Map(X, Y) \in SH$ (the stable homotopy category) for any $X, Y \in \mathcal{D}$ satisfying "the usual" properties and that the functors $i_*$ etc are similarly enriched. I then suspect the following:

Given $X, Y \in \mathcal{D}$, there is a distinguished triangle

$Map(X, Y) \xrightarrow{i_*i^*, j_!j^!} Map(i_*i^* X, i_*i^* Y) \oplus Map(j_! j^! X, j_! j^! Y) \xrightarrow{p-q} Map(i_*i^* X, j_!j^! Y[1]).$

Here $p$ is just composition with the map $w_Y$, and $q$ is composition with $w_X$ and using $Map(j_!j^! X, j_!j^! Y) = Map(j_!j^! X[1], j_!j^! Y[1])$. You may wish to view this as a homotopy (bi)cartesian square of spectra as usual.

My question is: is this result established somewhere I can cite? Or alternatively what is a good way of proving it?

Some ideas on a proof

The main observation is that $Map(j_!j^! X, i_*i^* Y) \simeq 0$ because $i^*j_! = 0$ follows from the axioms. This begs to be combined with the gluing triangles, of course. From this I have tried to proceed in a few ways:

Using functoriality of mapping spectra and playing around with the octahedral axiom, I can show that there is a distinguished triangle with the vertices as claimed, but I do not know that the maps are correct (one of the inclusions and one of the projections are ok, the others I cannot control). As usual with the octahedral axiom the argument is a bit space intensive but I can reproduce it if you want.

The result is quite transparent if we pretend that we are allowed to build complexes out of the mapping spectra; one may then directly show that $Map(X, Y)$ is quasi-isomorphic to the complex the triangle predicts. I think this can be made rigorous using dg categories and twisted complexes.

Lurie explains in appendix A.8 of higher algebra that (more or less) all recollements come from a "left exact correspondence" between categories. It appears to me that one may be able to establish the desired homotopy cartesian square from this description; unfortunately I am not sufficiently skilled in the manipulation of infinity categories to be sure.

Answer 1

I'm going to do a proof assuming we are in a stable $\infty$-category (I'm pretty sure this is almost equivalent to your "sufficiently rich" situation anyway). In your case $F=j_!j^!$ and $G=i_*i^*$.

Theorem: Let $C$ be a stable $\infty$-category, $F,G$ exact endofunctors of $C$ such that

• There's a fiber sequence of functors $F\to 1_C\to G$;
• For any $X,Y\in C$ the mapping spectrum $Map(FX,GY)$ is contractible.

Then the square $$\require{AMScd} \begin{CD} Map(X,Y) @>>> Map(GX,GY);\\ @VVV @VVV \\ Map(FX,FY) @>>> Map(GX,\Sigma FY); \end{CD}$$ is cartesian.

(this implies your fiber sequence by the standard argument turning a cartesian square in a Mayer-Vietoris long exact sequence).

Lemma 1: For any $X,Y$ the map $Map(FX,FY)\to Map(FX,Y)$ is an equivalence.

Proof: We have a fiber sequence $Map(FX,FY)\to Map(FX,Y)\to Map(FX,GY)$ by appling the exact functor $Map(FX,-)$ to $F\to 1\to G$ and the latter term is contractible. $\square$

Lemma 2: There is a fiber sequence $$Map(GX,Y)\to Map(X,Y)\to Map(FX,FY)$$

Proof: We have a fiber sequence $Map(GX,Y)\to Map(X,Y)\to Map(FX,Y)$ by applying the exact functor $Map(-,Y)$ to $FX\to X \to GX$ and Lemma 1 concludes. $\square$

Proof of the main result: By lemma 2 the fiber of the left column is $Map(GX,Y)$. By applying the exact functor $Map(GX,-)$ to $1\to G\to \Sigma F$ we have that $Map(GX,Y)$ is also the fiber of the right column. Hence the total fiber of the diagram is contractible, that is the diagram is cartesian. $\square$