Let $M$ be an $n$-dimensional orientable non-compact manifold.

Is there an isomorphism as follows, and if so how can we construct it? (Or can you provide a reference?) $$ H^{n-i}_{\operatorname{dR},c}(M, \mathbb R) \cong H_i(M,\mathbb R). $$ On the left hand side we have de Rham cohomology with compact support and on the right hand side we have singular homology.

According to [1], Poincaré duality can be stated as saying that the integration pairing between compactly supported forms and forms induces an isomorphism: $$ H^i_{\operatorname{dR}}(M,\mathbb R)\cong \left(H^{n-i}_{\operatorname{dR},c}(M, \mathbb R)\right)^\vee. $$ Where $\bullet^\vee$ denotes the vector space dual. Now, note that $H^{n-i}_c\not\cong (H^i)^\vee$ in general, because in our situation the homology groups might be infinite dimensional. The de Rham theorem says that integration gives an isomorphism witht he dual of singular homology: $$ H^i_{\operatorname{dR}}(M,\mathbb R)\cong \left(H_{i}(M, \mathbb R)\right)^\vee. $$ Now, knowing that two (infinite dimensional) vector spaces have the same dual doesn't seem all that helpful. I don't even see how integration can give a map $H^{n-i}_c\to H_i$.

Another attempt would be to try to follow the proof that appears in [1] and [2]. You could try to "induct" by showing that if Poincaré duality holds in two open sets $U$ abd $V$ then it holds in the union. But the five-lemma together with the Mayer-Vietoris exact sequences aren't enough to construct a map, much less a canonical one.

I'd be happy with an answer that uses Verdier duality (bonus points for an answer with coefficients in a local system), but then the trouble is that books that talk about sheaves don't talk about homology (or they define it as the cohomology with compact support of the dual local system, in which case this is tautology). Then the question becomes:

For a local system $\mathcal L$ on $M$, are the following quasiisomorphic? $$Rp_!(\mathcal L) \cong C_\bullet(M,\mathcal L):= C_\bullet(\widetilde M, \mathbb R)\otimes_{\mathbb R[\pi_1(M)]} \mathcal L_p.$$ Here $p:M\to *$ is the map to a point, $\widetilde M$ is the universal cover, and the fundamental group acts on $\widetilde M$ by deck transformations and on the stalk $\mathcal L_p$ by the monodromy of $\mathcal L$.

I guess this would be a corollary of the question

Is there a soft complex of sheaves that resolves $\mathcal L$, and whose compactly supported sections form a complex quasiisomorphic to $C_\bullet(M, \mathcal L)$?

Update: I've been looking at Glen Bredon's book [3], and Theorem V.9.2. is promising in that it relates homology and compactly supported cohomology. However, the definition of "sheaf homology" in that book doesn't seem to be related to singular homology as far as I can tell. In Chapter VI there is a relation between singular homology and Čech homology, but somehow not between these two and sheaf homology.

[1]: Greub, Werner; Halperin, Stephen; Vanstone, Ray, Connections, curvature, and cohomology. Vol. I: De Rham cohomology of manifolds and vector bundles, Pure and Applied Mathematics, 47. New York-London: Academic Press. XIX, 443 p. $ 31.00 (1972). ZBL0322.58001.

[2]: Hatcher, Allen, Algebraic topology, Cambridge: Cambridge University Press (ISBN 0-521-79540-0/pbk). xii, 544 p. (2002). ZBL1044.55001.

[3]: Bredon, Glen E., Sheaf theory., Graduate Texts in Mathematics. 170. New York, NY: Springer. xi, 502 p. (1997). ZBL0874.55001.

  • 2
    $\begingroup$ If the manifold $M$ has a good cover, then its (compact) cohomology is finite dimensional (see Bott-Tu, Differential forms in algebraic topology, Cha. 1, prop 5.3.1 and 5.3.2). Having a good cover is not a strong assumption (unless you are interested in a very special class of manifolds), for example a smooth manifold always has a good cover. $\endgroup$
    – AG learner
    Commented Mar 28, 2020 at 5:28
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    $\begingroup$ Thanks. Unfortunately, it's not true that smooth manifolds have finite dimensional homology or cohomology. In Bott-Tu, they say that this holds if a manifold has a finite good cover. Every manifold indeed has a good cover, but not necessarily finite. Take for example a connected infinite sheeted covering of a genus 2 surface or of a plane with two punctures. These are smooth, but their first (co)homology is infinite dimensional, so they can't have a finite good cover. $\endgroup$
    – Moisés
    Commented Mar 28, 2020 at 5:38
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    $\begingroup$ By "infinite sheeted" I guess I meant "with deck group Z". Anyway, an easier example is just the plane with the set $\mathbb Z\times 0$ removed, which has the homotopy type of a bouquet of countably many circles. $\endgroup$
    – Moisés
    Commented Mar 28, 2020 at 6:42
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    $\begingroup$ I think one can at least get a map one way by using Thom's theorem to represent a class in $H_i$ by a multiple of a submanifold $N$ (I think it works for non-compact $M$, does it?), then thake the Thom class of the normal bundle (compactly supported near the zero section, so transferable to neighborhood of $N$; in fact in Thom's theorem the normal bundle can be taken to be trivial). I don't know how to show that this is an isomorphism. $\endgroup$
    – Max
    Commented Apr 2, 2020 at 0:46

2 Answers 2


This is not a conclusive answer, but here's how I would approach this.

Let $\Delta_n(M)$ be the Abelian group of singular $n$-chains and $\Delta^n(M;\mathbb{R})=\mathrm{Hom}_\mathbb{Z}(\Delta_n(M),\mathbb{R})$ the $\mathbb R$-valued signular $n$-cochains. There are two subcomplexes that are relevant to the question: the complex $\Delta_*^\infty(M)$ of smooth singular chains (as explained in Bredon's book, for example), and the complex $\Delta_c^*(M)$ of compactly supported cochains, i.e. singular cochains that vanish on all chains with image outside of a compact set (which depends on the cochain).

Now, without having a reference or a proof, I would bet some money that the inclusion $\Delta_*^\infty(M)\hookrightarrow\Delta_*(M)$ is a chain homotopy equivalence. This should, in turn, dualize to chain homotopy equivalences $$\Delta^*(M;\mathbb R)\to\Delta^*_\infty(M;\mathbb R) \quad\text{and}\quad \Delta^*_c(M;\mathbb R)\to \Delta^*_{\infty,c}(M)$$ where $\Delta^n_\infty = \mathrm{Hom}(\Delta_n^\infty,\mathbb R)$ and $\Delta^n_{\infty,c}$ is the compactly supported analogue.

Next, integration gives rise to chain maps $$\Psi\colon \Omega^*(M)\to \Delta_\infty^*(M) \quad\text{and}\quad \Psi_c\colon \Omega^*_c(M)\to \Delta_{\infty,c}^*(M).$$ Bredon proves that $\Psi$ induces an isomorphism on cohomology and it should be possible to adapt the proof to show the same for $\Psi_c$.

Finally, the cohomology of $\Delta_c^*(M)$ is known a singular cohomology with compact supports, denoted by $H^*_c(M;\mathbb R)$. If $M$ has an orientation, then Poincaré duality gives isomorphisms $$H^{n-i}_c(M;\mathbb R)\cong H_{i}(M;\mathbb R)$$ with singular homology on the right hand side. And if everything above goes through as claimed, then the left hand side is isomorphic to compactly supported de Rham cohomology $H^{n-i}_{dR,c}(M)$.

I'll try to find some references later. Other duties are calling.


It turns out that Glen Bredon's book does answer this question, but to me it seems very difficult to understand the map just from what's in the book. Theorem V.9.2. states a version of Poincaré duality for any sheaf. Further, Theorem V.12.21 states that there is an isomorphism between "sheaf homology" as defined in Chapter V and singular homology defined using chains. However, understanding this isomorphism seems to require understanding sheaf homology and right now I'm not up for the task.

However, Bredon sketches another answer in IV.2.9. (together with exercises I-12 and II-32), which I'm going to write in more detail here, as I have two issues with it. First, it only proves the duality for constant coefficients, because the general version with coefficients in a local system is a consequence of V.9.2 and V.12.21, which as I said above I don't really understand. Secondly, the sentence in IV.2.9. at the top of page 207 ("If X is a manifold, it is clear that...") is not clear to me.$\newcommand{\cL}{\mathcal L}\newcommand{\qiso}{\overset{\text{qiso}}\cong}\newcommand{\Z}{\mathbb Z}$

Let $M$ be an $n$-dimensional orientable (non-compact) manifold, let $\cL$ be a local system on $M$ and let $p:M\to \text{pt}$ be the map to a point. Our goal is to construct a soft resolution $S^0\to \cdots \to S^n$ of $\cL$ with the property that $p_!(S^\bullet) \cong C_{n-\bullet}(M;\cL)$ is the complex of $\cL$-valued singular chains.

Let us first see why this is what we need. By the softness if $S^\bullet$, this would imply that $$ Rp_!\cL \overset{\text{qiso}}\cong Rp_!(S^\bullet) \qiso p_!(S^\bullet) \cong C_{n-\bullet}(M;\cL). $$ So $H^i_c(M;\cL)\cong H_{n-i}(M;\cL)$. On the other hand, $\cL$ is resolved by the de Rham complex $\Omega^\bullet(\cL)$, which is also soft. Therefore, $$ Rp_!\cL\overset{\text{qiso}}\cong p_!\Omega^\bullet(\cL). $$ So $H^i_c(M;\cL)\cong H^i_{c,\text{dR}}(M;\cL)$. These two isomorphisms together provide the answer.

Here is how we construct the resolution. $S^{n-i}$ will turn out to be the sheafification of the presheaf that assigns to an open set $U\subset M$ the space $C_i(M,M\setminus U;\cL)\cong \frac{C_i(M;\cL)}{C_i(M\setminus U;\cL)}$. Note that if $V\subseteq U$, there is a map $C_i(M,M\setminus U;\cL)\to C_i(M,M\setminus V;\cL)$.

$S^{n-i}$ is given as follows. We denote $$ C_i^{\Pi}(X;\cL)=\prod_\phi\Gamma(\Delta_i;\phi^*\cL)$$ Where $\phi:\Delta_i\to X$ ranges over all the singular simplices (so we are using infinite chains as opposed to usual finite chains). For an open set $U\subset M$, we define $ C_i^U(X;\cL) \subseteq C_i^{\Pi}(X;\cL) $ as the subgroup of chains $(s_\phi)_{\phi:\Delta_i\to X}$ with the property that every $p\in U$ has a neighborhood intersecting only finitely many chains $\phi(\Delta_i)$ for which $s_\phi\neq 0$ (so if $X=U$ this corresponds to the Borel-Moore chains). We define $$ S^{n-i}(U) = C_i^U(M|U;\cL):=\frac{C_i^U(M;\cL)}{C_i^U(M\setminus U;\cL)} . $$ Throughout the coefficients will always be $\cL$, so we will ommit it from the notation.

The first claim is that $S^{n-i}$ is a sheaf. Suppose we are given a covering $U = \bigcup V_j$ and compatible sections $(s^j_\phi)_{\phi}\in S_i(V_j)$. The compatibility means that if $\phi\cap V_j \cap V_{j'}\neq \emptyset$, then $s^j_\phi = s^{j'}_\phi$. Therefore, there is a unique chain $(s_\phi)_{\phi\cap U\neq \emptyset}$, given by $s_\phi = s_\phi^j$ for any $j$ for which $\phi\cap V_j\neq \emptyset$ (it is unique modulo chains contained in $M\setminus U$, as desired). It remains to check that this chain belongs to $C_i^U(M)$: for any $p\in U$, $p$ must be contained in some $V_j$, and the finiteness property follows from $(s_\phi)|_{V_j} = (s_\phi^j)$.

Note that this is not exactly the group of Borel-Moore chains, as the chains in $S_i(U)$ are allowed to accumulate on the boundary of $U$, unlike the case of the Borel-Moore group.

Next, we verify that $S^{n-i}$ is a (c-)soft sheaf: it is easy to see that sections on a compact set $K$ are simply given by $S^{n-i}(K) = C_i(M,M\setminus K)$, which has a surjective map from $S_i(M) = C_i^{BM}(M)$. From the definition, we can check directly that the compactly supported global sections of $S_i$ are isomorphic to $C_i(M)$.

Finally, we want to show that the complex $S^0\to S^1\to\cdots \to S^n$ resolves $\cL$ (here if $M$ was non-orientable we would have to twist by the orientation sheaf). Let $U$ be a ball around a point in $M$. First of all, we claim that $$ C_i^U(M|U) \cong \lim_{\substack{\gets \\K \text{ compact}\\ K \subset U}} C_i^U(M|K). $$ There is a map going from left to right. Suppose a chain $(s_\phi)$ maps to $0$. Then its support does not intersect any compact $K\subset U$, i.e. it's supported on $M\setminus U$. This shows the map is injective. To see it is surjective, an element on the right hand side is a compatible sequence of chains $(s_\phi^K)_\phi$, which glue to a chain $(s_\phi)\in C_i^U(M|U)$ by the same argument that we used to show that $S_i$ is a sheaf.

Next, we check that excision works in this situation. Give two sets $A,B$ with $\overline A\cap \overline B = \emptyset$, we have that the inclusions induce a homotopy equivalence $C_\bullet^U((X\setminus A)|B) \cong C_\bullet^U(X|B)$. We can verify this by looking at the proof in Hatcher [2]: it shows that barycentric subdivision provides a homotopy inverse for the chain map. Barycentric subdivision is defined at the level of chains, and can be extended to infinite chains with the same formulas.

Now, applying excision to the sets $M\setminus U$ and $K$ tells us that there is a homotopy equivalence $ C_\bullet^U(M|K) \cong C_\bullet^U(U|K)$, and further shrinking $U$ to a compact neighborhood $K_2$ of $K$ in $U$ we have that we can work with usual chains: $$ C_\bullet^U(U|K) \qiso C_\bullet^U(K_2|K) = C_\bullet(K_2|K) \qiso C_\bullet(U|K). $$ As far as I know, we should be careful that the inverse limit of the homotopy equivalences we have constructed stays a quasiisomorphism after taking the limit. Consider the cone of the map $C_\bullet(U|K)\to C_\bullet^U(U|K)$, let's call it $N_\bullet^K$. For every $K$, its homology is trivial, as it is the cone of a quasiisomorphism. Since all the maps involved in the limit are surjective, Stacks Project 0918 suffices to show that $\lim_{\gets,K} N_\bullet^K$ has vanishing homology, and therefore the limit map $\lim_\gets C_\bullet(U|K)\to \lim_\gets C_\bullet^U(U|K)$ is a quasiisomorphism, as its cone's homology vanishes. We have shown that the inclusion maps induce a quasiisomorphism: $$ \lim_{\substack{\gets \\K \text{ compact}\\ K \subset U}} C_\bullet(U|K) \overset{\sim}{\to} S^{n-\bullet}(U) = C_\bullet^U(M|U). $$ Now, if $U$ is a ball and $K\subseteq K'$ are closed balls (possibly of radius $0$), the map $C_i(U|K')\to C_i(U|K)$ is a surjective quasiisomorphism, since homology is homotopy invariant. Any compact set is contained in a closed ball, so the limit can be computed using just closed balls. Another application of Stacks 0918 shows that for any point $p\in U$, there is a quasiisomorphism $$ \lim_{\substack{\gets \\K \text{ compact}\\ K \subset U}} C_\bullet(U|K) \overset{\sim}\to C_\bullet(U|p). $$ Now, $U$ is a ball, so we can canonically identify $\cL|_U$ with the constant sheaf $\underline{\cL_p}$. Finally, Hatcher 3.3 tells us that $H_i(U|p,\underline{\cL_p})\cong \widetilde H_{i-1}(S^{n-1},\cL_p)$. Since $\cL_p$ is torsion-free (it is a vector space over $\mathbb R$ or $\mathbb C$), the universal coefficient theorem says that $H_{i-1}(S^{n-1},\cL_p)\cong H_{i-1}(S^{n-1},\Z)\otimes_\Z \cL_p$. The term $H_{i-1}(S^{n-1},\Z)$ is the orientation sheaf, as a section corresponds to an orientation of $S^{n-1}$ at every point. Further, if we go around a loop the homology class will change according to the monodromy of $\cL_p$, and possibly by a sign if $M$ is not orientable and the loop reverses the orientation. This shows that $S_i$ is a resolution of $\cL$ (in the orientable case), and we are done. Tracing the above proof, the map $\cL\to S^0$ is given by mapping a section $s$ to a chain $\sum \phi_i^*(s)$, where $\phi_i:\Delta_n\to M$ are simplices such that $\sum \phi_i = [M]$, the (Borel-Moore) fundamental class of $M$.


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