In the paper "Resolutions of unbounded complexes" (Compositio Math., vol. 65, no. 2, pp. 121-154) N. Spaltenstein generalizes the 6 functor formalism to unbounded complexes of sheaves over ringed spaces. The properties which only involve the direct and inverse image functors, Hom and the tensor product are proved in full generality (Theorem A). For statements that also involve the direct and inverse images with compact support (Theorem B) one needs the spaces to be locally compact and Hausdorff, which is not surprising.


However, in this case Spaltenstein also makes the additional assumption that the kernel of the differential in a (possibly unbounded) complex of c-soft sheaves is c-soft in each degree (Condition ($\ast$) on p. 122, which is used in Theorem B). He remarks that this assumption is satisfied for a space $X$ if it is locally finite dimensional, meaning that for every $x\in X$ there is an $n\in\mathbb{Z}$ such that $x\in X$ has an arbitrarily small neighborhood $U$ with the property $H^{>n}\_c(U,S)=0$ for every sheaf $S$ on $U$. So in particular, all locally finite polyhedra have property ($\ast$).


Condition ($\ast$) is needed to construct a class $\mathfrak{S}$ of complexes of sheaves that has the properties listed on p. 150. Namely, if the condition holds, then one can take $\mathfrak{S}=$ the class of complexes of degree-wise c-soft sheaves.


From reading the paper one gets the impression that the condition is a technical shortcut which may not be absolutely necessary. Also, the paper appeared in 1988, which is quite a while ago, so one wonders whether some of the technical issues have since been settled. Namely,


(1) Can one construct a class $\mathfrak{S}$ of complexes that has properties (a)-(f) on p. 150 without using Condition ($\ast$)?


(2) Are there compact Hausdorff spaces that do not satisfy Condition ($\ast$)? Also, is there a wider class of locally compact Hausdorff spaces than locally finite dimensional for which Condition ($\ast$) holds?



Resolutions of unbounded complexes: Condition ($\ast$) in Spaltenstein's paper
Here is a variant of an example due to Lurie (as far as I can tell) [HTT, Counterexample 6.5.4.2] showing that the proper base change theorem [Spaltenstein, Proposition 6.20] does *not* hold for unbounded complexes, even on compact Hausdorff spaces. In particular, as condition 6.14(2) can be replaced by 6.14(a)–(f) or condition (\*), the example shows that a class $\mathfrak S$ with these conditions does not always exist.


In fact, Lurie uses this as one of the motivations why the natural notion of $\infty$-topos is defined in terms of sheaves instead of hypersheaves; see [HTT, Warning 6.5.4.1]. I'm not entirely sure what this translates to in the language of derived categories, but I suspect the "correct" category will somehow look more like "sheaves of spectra" than like chain complexes in $\operatorname{Sh}(X,\mathbf Z)$.


For self-containedness, let me translate the example to the language of Spaltenstein. It turns out that this is a little more involved, roughly because $\operatorname{Hom}(\mathbf Z^{(S)},\mathbf Z^{(T)})$ is harder to compute than $\operatorname{Map}(S,T)$ for sets $S$ and $T$; see the remark after the example.


**Example.** Let $Q = [0,1]^{\mathbf N}$ be the Hilbert cube (where $\mathbf N = \{0,1,\ldots\}$). For $i \in \mathbf N$, write $Q\_i = [0,1]^{\mathbf N\_{> i}}$, so that $Q = [0,1]^{\{0,\ldots,i\}} \times Q\_i = [0,1]^{i+1} \times Q\_i$. For $i \in \mathbf N$, define opens in $X = Q \times [0,1]$ by
\begin{align\*}
U\_i &= (0,1)^i \times [0,1)\times Q\_i \times [0, 2^{-i}),\\
V\_i &= (0,1)^i \times (0,1] \times Q\_i \times [0, 2^{-i}),\\
\end{align\*}
and write $U\_{i,0} = U\_i \cap (Q \times \{0\})$ and $V\_{i,0} = V\_i \cap (Q \times \{0\})$. Then we have $U\_i \cup V\_i \subseteq U\_{i-1} \cap V\_{i-1}$ for all $i \geq 1$, and this induces an equality
$$U\_{i,0} \cup V\_{i,0} = U\_{i-1,0} \cap V\_{i-1,0}.\tag{1}\label{1}$$
Consider the complex $C^\bullet \in \operatorname{Ch}^{\leq 0}(X,\mathbf Z)$ given by
$$\cdots \to \mathbf Z\_{U\_i} \oplus \mathbf Z\_{V\_i} \to \cdots \to \mathbf Z\_{U\_1} \oplus \mathbf Z\_{V\_1} \to \mathbf Z\_{U\_0} \oplus \mathbf Z\_{V\_0},$$
where the differentials are all given by the matrix $\big(\begin{smallmatrix}1 & 1 \\ -1 & -1\end{smallmatrix}\big)$. Consider the base change square
$$\begin{array}{ccc}Q \times \{0\} & \stackrel{q'}\to & X \\ \!\!\!\!\!{\scriptsize f'}\!\downarrow & & \downarrow\!{\scriptsize f}\!\!\! \\ \{0\} & \stackrel q\to & [0,1].\!\end{array}$$
We claim that the natural map $q^\*Rf\_! C^\bullet \to Rf'\_!q'^\* C^\bullet$ is not an isomorphism in $D(\{0\},\mathbf Z) = D(\mathbf Z)$. In fact, we will show that the left hand side is $\mathbf Z[-1]$ and the right hand side $\mathbf Z[0]$, so the map is zero as $\operatorname{Hom}\_{D(\mathbf Z)}(\mathbf Z[-1],\mathbf Z) = \operatorname{Ext}^1(\mathbf Z,\mathbf Z) = 0$.


For the right hand side, restricting to $Q \times \{0\}$ gives the complex
$$\cdots \to \mathbf Z\_{U\_{i,0}} \oplus \mathbf Z\_{V\_{i,0}} \to \cdots \to \mathbf Z\_{U\_{1,0}} \oplus \mathbf Z\_{V\_{1,0}} \to \mathbf Z\_{U\_{0,0}} \oplus \mathbf Z\_{V\_{0,0}},$$
which is acyclic in negative degree by \eqref{1} and the Mayer–Vietoris sequences
$$0 \to \mathbf Z\_{U \cap V} \stackrel{\big(\begin{smallmatrix}1 \\ -1\end{smallmatrix}\big)}\longrightarrow \mathbf Z\_U \oplus \mathbf Z\_V \stackrel{(1\ 1)}\longrightarrow \mathbf Z\_{U \cup V} \to 0.$$
Since $U\_{0,0} \cup V\_{0,0} = Q \times \{0\}$, the Mayer–Vietoris sequence then shows that the natural quotient
$$q'^\*C^\bullet \to \mathbf Z\_{Q \times \{0\}}[0]$$
is a quasi-isomorphism. Thus, $Rf'\_!q'^\* C^\bullet = Rf'\_!\mathbf Z[0] = \mathbf Z[0]$ since $Q$ is compact and contractible.


For the left hand side, we have to do a little work to show that $C^\bullet$ is Postnikov complete, which is the only way I know how to compute cohomology of complexes that are unbounded below on an infinite-dimensional space. (Please let me know if you know a different method!)


For each finite set $I \subseteq \mathbf N \cup \{\infty\}$, write $\pi\_I \colon X \to [0,1]^I$ for the projection (where $\infty$ is the separate factor $[0,1]$ in $X = Q \times [0,1]$). Consider the set $\mathfrak P$ of sets of the form $\pi\_I^{-1}(A)$, where $I \subseteq \mathbf N$ is finite and $A = \prod\_{i \in I} A\_i$ is a product of intervals (open, half-open, or closed). Note that $U\_i$, $V\_i$, $U\_i \cup V\_i$, and $U\_i \cap V\_i$ are in $\mathfrak P$ for all $i$, as are the sets
$$W\_i = \big(U\_i \cap V\_i\big)\setminus\big(U\_{i+1} \cup V\_{i+1}\big) = (0,1)^{i+1} \times Q\_i \times [2^{-i-1}, 2^{-i}).$$
Since the intersection of two intervals is an interval, the set $\mathfrak P$ is closed under finite intersections. For $x \in X$, write $\mathfrak P\_x = \{A \in \mathfrak P\ |\ x \in A^\circ\}$, and note that this forms a neighbourhood basis of $x$ (it will be convenient *not* to restrict solely to open neighbourhoods). Write $\mathfrak B = \operatorname{Open}(X) \cap \mathfrak P$ and $\mathfrak B\_x = \mathfrak B \cap \mathfrak P\_x$; these form a basis for the topology and for the open neighbourhoods of $x$ respectively. Finally, note that if $A \in \mathfrak P$, then $\bar A \in \mathfrak P$.


**Lemma.** *If $A \in \mathfrak P$ and $x \in X$, then $H^i(V,\mathbf Z\_A) = 0$ for all $i > 0$ and $V \in \mathfrak P\_x$ sufficiently small.*


In particular, the opens $\mathfrak B$ and the sheaves $\mathbf Z\_A$ for $A \in \mathfrak P$ satisfy condition 3.12(1) in [Spaltenstein].


*Proof.* Let $B = \bar A$ and $D = B \setminus A$. If $A = \pi\_I^{-1}\big(\prod\_{i\in I} A\_i\big)$ with $\bar A\_i = [a\_i,b\_i]$, then
$$D \subseteq \partial A = \pi\_I^{-1}\big(\prod\_{i \in I} \{a\_i,b\_i\}\big).$$
Call the *faces* of $D$ the subsets of the form $\pi\_i^{-1}(\{a\_i\})$ or $\pi\_i^{-1}(\{b\_i\})$. Consider those $V \in \mathfrak P\_x$ such that $V \cap \partial A$ only meets the faces containing $x$; concretely this can be accomplished by taking $V$ contained in $\pi\_I^{-1}\big(\prod\_{i \in I}[x\_i-\varepsilon,x\_i+\varepsilon]\big)$ for $\varepsilon$ such that $\varepsilon < \lvert x\_i-a\_i\rvert$ whenever $x\_i \neq a\_i$, and likewise for $b\_i$. This means that
$$V \cap D = V \cap \bigcup\_{i \in F} \pi\_i^{-1}(\{x\_i\}) \qquad \text{where } F = \{i \in I\ |\ x\_i\in\{a\_i,b\_i\}\}.$$
Thus, $V \cap D$ is either empty (if $F = \varnothing$) or contractible. Since the locally closed immersion $A\cap V \to V$ factors via the closed immersion $B \cap V \to V$, we get
$$R\Gamma(V,\mathbf Z\_A) = R\Gamma(V,\mathbf Z\_{A \cap V}) = R\Gamma(B \cap V,\mathbf Z\_{A \cap V}).$$
If $F = \varnothing$, then $A \cap V = B \cap V$, so we conclude that $R\Gamma(V,\mathbf Z\_A) = \mathbf Z[0]$ since $B \cap V$ is contractible. Likewise, if $F \neq \varnothing$, then $V \cap D$ is contractible, so the map $R\Gamma(B \cap V,\mathbf Z\_{B \cap V}) \to R\Gamma(B \cap V,\mathbf Z\_{D \cap V})$ is an isomorphism, so we conclude $R\Gamma(B \cap V,\mathbf Z\_{A \cap V}) = 0$. $\square$


Going back to the complex $C^\bullet$ above, we have $\mathscr H^{-i}(C^\bullet) \cong \mathbf Z\_{W\_i}$ by Mayer–Vietoris, since the kernel of $d^{-i}$ is $\mathbf Z\_{U\_i \cap V\_i}$ and the image of $d^{-i-1}$ is $\mathbf Z\_{U\_{i+1}\cup V\_{i+1}}$. Thus the lemma above and [Spaltenstein, Proposition 3.13] (see also [Stacks, Tag [0D63](https://stacks.math.columbia.edu/tag/0D63)]) imply that
$$C^\bullet \to \underset{\substack{\longleftarrow \\ n}}{\operatorname{holim}} \tau\_{\geq -n}C^\bullet$$
is an isomorphism in $D(X,\mathbf Z)$. Then [Stacks, Tag [0D60](https://stacks.math.columbia.edu/tag/0D60)] shows that
$$R\Gamma\big(Q \times [0,a),C^\bullet\big) = \underset{\substack{\longleftarrow \\ n}}{\operatorname{holim}} R\Gamma\big(Q \times [0,a), \tau\_{\geq -n} C^\bullet\big),$$
so we have to compute $R\Gamma(Q \times [0,a),\tau\_{\geq -n}C^\bullet)$. For $n < 0$ we have $\tau\_{\geq -n} C^\bullet = 0$. By [Stacks, Tag [08J5](https://stacks.math.columbia.edu/tag/08J5)] we have distinguished triangles
$$\mathscr H^{-i}(C^\bullet)[i] \to \tau\_{\geq -i} C^\bullet \to \tau\_{\geq -i+1} C^\bullet.\tag{2}\label{2}$$
Recall that $\mathscr H^{-i}(C^\bullet) = \mathbf Z\_{W\_i}$. Thus the same computation as in the lemma (plus the proper base change theorem and the computation of $R\Gamma\_c\big((0,1)^{i+1},\mathbf Z\big)$) gives
$$R\Gamma\big(Q \times [0,a),\mathbf Z\_{W\_i}\big) = \begin{cases}0, & a > 2^{-i},\\ \mathbf Z[-i-1], & 2^{-i-1} < a \leq 2^{-i}, \\ 0, & a \leq 2^{-i-1}.\end{cases}$$
Indeed, in the first case, the closed interval $[2^{-i-1},2^{-i}]$ is contained in $[0,a)$, so again the inclusion
$$(0,1)^{i+1} \times Q\_i \times \{2^{-i}\} \to (0,1)^{i+1} \times Q\_i \times [2^{-i+1},2^{-i}]$$
induces an isomorphism on compactly supported cohomology. In the second case, $W\_i \cap (Q \times [0,a))$ is closed and contractible in $Q \times [0,a)$, and in the third case it is empty.


Given $a \in (0,1]$, there is a unique $i \in \mathbf N$ with $2^{-i-1} < a \leq 2^{-i}$. From \eqref{2} we thus conclude that
$$R\Gamma\big(Q \times [0,a),\tau\_{\geq -n} C^\bullet\big) = \mathbf Z[-1]$$
as soon as $n \geq i$. Thus, the cohomology of $\tau\_{\geq -n}C^\bullet$ is eventually constant for all $n$ and $a$, and we conclude $R\Gamma(Q \times [0,a),C^\bullet) = \mathbf Z[-1]$. Since $R^if\_!C^\bullet = R^if\_\*C^\bullet$ is the sheafification of $U \mapsto H^i(f^{-1}(U),C^\bullet)$ [Stacks, Tag [0D5X](https://stacks.math.columbia.edu/tag/0D5X)] and sheafification preserves stalks, we conclude that
$$q^\*Rf\_!C^\bullet = \mathbf Z[-1],$$
which has cohomology only in degree $1$, and does not agree with $Rf'\_!q'^\*C^\bullet$. $\square$


**Remark.** The example of [HTT, Counterexample 6.5.4.2] is much easier, because it works with sheaves of spaces instead of chain complexes of sheaves of abelian groups. The point is that the sheaf of sets $\pi\_0(\mathscr F)$ of Lurie's example has empty global sections, so $\mathscr F$ itself cannot have global sections. But in abelian categories, nonzero objects can of course map to the zero object, and the vanishing of $q^\*Rf\_\*C^\bullet$ was substantially more work. There might be shortcuts to the above argument using simplicial or homotopical techniques.


**Remark.** One can immediately see that the Hilbert cube is not locally finite-dimensional: opens around an 'interior' point (no coordinate is $0$ or $1$) have higher and higher cohomological dimension, already for compactly supported cohomology with constant coefficients.


On the other hand, checking (\*) directly in this case is a little unclear, but the failure of proper base change shows that it cannot be true. So Spaltenstein wasn't wrong to impose extra hypotheses!




---


**References.**


[HTT] J. Lurie, [*Higher topos theory*](http://dx.doi.org/10.1515/9781400830558). Annals of Mathematics Studies **170**. Princeton University Press, 2009.


[Spaltenstein] N. Spaltenstein, [*Resolutions of unbounded complexes*](http://www.numdam.org/item?id=CM_1988__65_2_121_0). Compos. Math. **65**.2, p. 121-154 (1988).


[Stacks] [The Stacks project](https://stacks.math.columbia.edu).


