In many reasonable six-functor formalisms, open and closed immersions satisfy the so-called recollement conditions. (This holds in all the "constructible" formalisms. For example, in the analytic setting, in the Γ©tale setting, for D-modules, etc.) Let me precise what I mean by that.

(Recollement) Let $i\colon Z\hookrightarrow X$ be a closed immersion and $j\colon U\hookrightarrow X$ be its complementary open immersion. Recall that $i$ is proper (and so $i_!\cong i_*$) and $j$ is Γ©tale (and so $j^!\cong j^*$). We have that:

  • $i^!j_*=0$ (so, by adjunction, $i^*j_!=0$ and $j^*i_*=0$);
  • the adjunction maps $i_! i^! M \to M \to j_* j^* M$ and $j_! j^! M\to M^\bullet\to i_* i^* M$ give rise to distinguished triangles;
  • the adjunction maps $i^*i_*\to\operatorname{id}\to i^!i_!$ and $j^*j_*\to\operatorname{id}\to j^!j_!$ are all isomorphisms.

Now, I would hope that this (along with the usual machinery, such as proper / smooth base change, the projection formula, etc...) implies a Mayer-Vietoris-style result. Precisely, if $U_1,U_2$ are open subsets of $X$, I hope to have a distinguished triangle $$j_{U_1\cap U_2,!}j_{U_1\cap U_2}^*M\to j_{U_1,!}j_{U_1}^* M\oplus j_{U_2,!}j_{U_2}^* M \to j_{U_1\cup U_2,!}j_{U_1\cup U_2}^*M,$$ where all the $j$'s are the natural inclusions of the open sets into $X$.

Idea of proof: consider the following diagram, in which every morphism is an open immersion and in which the large square is both cartesian and cocartesian.

By proper base change and the commutativity of the diagram above, it suffices to construct a distinguished triangle

$$ b_!b^*c_! c^*N\to b_! b^* N\oplus c_! c^* N\to N. $$

(Put $N=j_{U_1\cup U_2}^* M$ and apply $j_{U_1\cup U_2,!}$ in the triangle above.) Since $b$ and $c$ are open immersions, we have that $b^*\cong b^!$ and $c^*\cong c^!$. In particular, we have adjunction maps

$$ \varphi:b_!b^*\to \operatorname{id}\qquad\text{and} \qquad \psi:c_!c^*\to \operatorname{id}. $$

We define the map $b_!b^*c_! c^*N\to b_! b^* N\oplus c_! c^* N$ to be $\begin{pmatrix}b_!b^*\psi_{N}\\ \varphi_{c_!c^+N^\bullet}\end{pmatrix}$ and the map $b_! b^* N\oplus c_! c^* N\to N$ to be $(\varphi_{N},-\psi_{N})$.

Now, does this give rise to the desired distinguished triangle? If so, how?

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    $\begingroup$ Yes it works. You only need recollement (no need for fancier smooth and/or proper base change formulas). What you want is a consequence of the octahedral axiom (if you work with stable $\infty$-categories, this is nicer, you just need to know that a square in a stable category is a pullback iff two parallel maps in it have isomorphic fibers), starting from the two cofiber sequences expressing the recollement of $U_1$ and $(U_1\cup U_2) - U_1$ and the recollement of $U_1\cap U_2$ and $U_2 - (U_1\cap U_2)$. $\endgroup$ Mar 29 at 19:02
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    $\begingroup$ The trick consists to observe what your data look like after apply $b^*$ or $c^*$. $\endgroup$ Mar 29 at 19:08
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    $\begingroup$ What you need it to consider the commutative square with first line 𝑏!π‘βˆ—π‘!π‘βˆ—π‘β†’π‘!π‘βˆ—π‘ and second line 𝑐!π‘βˆ—π‘β†’π‘. You want to prove that it is homotopy cocartesian, that is that, if you pick two parallel arrows, they have isomorphic cofiber. Since an object defined on the union of $U_1$ and $U_2$ is zero iff its restriction to each $U_i$ is zero, it suffices to check that your square is homotopy cocartesian after restrict to each $U_i$, and this is then obvious because you get squares in which two parallel maps are isomorphisms. $\endgroup$ Mar 30 at 9:53
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    $\begingroup$ @D.-C.Cisinski it seems to be subtle here to prove that the counit $c_!c^* \to \mathrm{id}$ (or $j_2$) is the unique one making the square homotopy cartesian. If one uses triangulated categories solely then I think it is due to the non-functoriality of the cone. $\endgroup$
    – Alexey Do
    Mar 30 at 12:21
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    $\begingroup$ @AlexeyDo All the triangles that appear in the proof are those coming from the recollement are unique (at least if we ask that they are functorially given, which the recollement structure ensures). $\endgroup$ Mar 30 at 13:52

1 Answer 1


As Cisinski commented, what you really want is a consequence of the octahedral axiom. However, there is a subtle point here, in [Neeman, Triangulated Categories, Definition 1.3.13] it is called a TR4'. In a triangulated category $\mathcal{T}$, a square $\require{AMScd}$ \begin{CD} Y @>{f}>> Z\\ @V{g}VV @VV{g'}V \\ Y' @>{f'}>> Z' \end{CD} is homotopy cartesian if there exists a distinguished triangle $$Y \overset{(g,-f)}{\longrightarrow} Y' \oplus Z \overset{f'+g'}{\longrightarrow} Z \longrightarrow +1$$ Now suppose we are given a diagram $\require{AMScd}$ \begin{CD} X @>{h}>> Y @>{f}>> Z @>{w}>>+1\\ @V{\mathrm{id}}VV @V{g}VV @VV{???}V @VVV \\ X @>{gh}>> Y' @>{f'}>> Z'@>{w'}>> +1 \end{CD} with $???$ unknown. Then it is possible to fill in $???$ an arrow such that the second square is homotopy cartesian [loc.cit, Lemma 1.4.3] and $$X \oplus Y \overset{\begin{pmatrix} -f & 0 \\ g & gh \\ \end{pmatrix}}{\longrightarrow} Y' \oplus Z \overset{\begin{pmatrix} -w & 0 \\ ??? & f' \\ \end{pmatrix}}{\longrightarrow} X[+1] \oplus Z' \longrightarrow +1$$ which turns out to be the sum of $$X \longrightarrow 0 \longrightarrow X[+1] \longrightarrow X[+1]$$ $$Y \longrightarrow Y' \oplus Z \longrightarrow Z' \longrightarrow Y[+1]$$ (sum of two triangulates is distinguished iff both are distinguished). Now for simplicity, let me assume that $X = U_1 \cup U_2$ and consider a diagram $\require{AMScd}$ \begin{CD} U_1 \cap U_2 @>{d}>> U_2\\ @V{a}VV @VV{j_2}V \\ U_1 @>{j_1}>> X \end{CD} and set $$i_1: X - U_1 \longrightarrow X, \ \ i_2: U_2 - (U_1 \cap U_2) \longrightarrow U_2.$$ and use the recollement of pairs $(i_1,j_1)$ and $(i_2,d)$ to get $$j_{2!}d_!d^*j_2^* \longrightarrow j_{2!}j_2^* \longrightarrow j_{2!}i_{2*}i_2^*j_2^* \longrightarrow +1$$ $$j_{1!}j_1^* \longrightarrow \mathrm{id} \longrightarrow i_{1*}i_1^* \longrightarrow +1.$$ Note that $$j_{2!}d_!d^*j_2^* = j_{12!}j^*_{12}, i_1 = j_2 \circ i_2.$$ In other words, the cones of $j_{12!}j^*_{12} \longrightarrow j_{2!}j_2^*$ and $j_{1!}j_1^* \longrightarrow \mathrm{id}$ coincide and then you rotate it and use above to see that it provides a Mayer-Vietoris sequence. $\require{AMScd}$ \begin{CD} \mathrm{Cone}[-1] @>>> j_{12!}j^*_{12} @>>> j_{2!}j_2^* @>>>+1\\ @V{\mathrm{id}}VV @VVV @VVV @VVV \\ \mathrm{Cone}[-1] @>>> j_{1!}j_1^* @>>> \mathrm{id} @>>> +1 \end{CD}

Caution: What I have proven so far is that there is a distinguished triangle $$j_{12!}j_{12}^* \longrightarrow j_{1!}j^*_1 \oplus j_{2!}j^*_2 \longrightarrow \mathrm{id} \longrightarrow +1,$$ where the $j_{2!}j^*_2 \longrightarrow \mathrm{id}$ maybe not the counit morphism. Apparently, it is due to the non-functoriality of the cone construction in triangulated categories (as a consequence, we may need our triangulated categories come from stable model categories, see this link, Proposition 4.3).

  • $\begingroup$ Why use derivators for such a simple question?? $\endgroup$ Mar 30 at 4:17
  • $\begingroup$ @Satan'sMinion edited $\endgroup$
    – Alexey Do
    Mar 30 at 11:12

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