Given a smooth manifold $M$ and a relatively compact exhaustion $M=\bigcup_{n\in\mathbb N} M_n$ with open and relatively compact $M_n\subseteq M_{n+1}$ (hence $M=\lim\limits_{\to}M_n$) do we always have $$H^k(M)=\lim\limits_{\leftarrow} H^k(M_n)?$$

This looks so natural that the answer should be known in which case I would like to get a reference.

EDIT (inspired by Weibel's book and an article of Milnor from 1962). For fixed $k$ the projective spectrum $\Omega^k(M_n)$ of $k$-forms on $M_n$ satisfies the (strict) Mittag-Leffler condition so that the derived functor ${\lim\limits_{\leftarrow}}^1 \Omega^k(M_n)$ vanishes. We thus get an exact sequence $0\to \Omega^\ast(M)\to \prod_n \Omega^\ast(M_n)\to \prod_n \Omega^\ast(M_n)\to 0$ of cochain complexes (the last map is the difference map $(\omega_n)_n\mapsto (\omega_n-\rho_{n+1}(\omega_{n+1}))_n$ with the restriction operator $\rho_{n+1}$). The long cohomology sequence thus yields an exact sequence $$\cdots \to H^k(M)\to \prod_nH^k(M_n) \to \prod_nH^k(M_n) \to H^{k+1}(M)\to \cdots.$$ From this one gets short exact sequences $$0\to {\lim\limits_{\leftarrow}}^1 H^{k-1}(M_n) \to H^k(M)\to \lim\limits_{\leftarrow}H^k(M_n)\to 0.$$ It seems to me that the spectrum $H^0(M_n)$ of the spaces of locally constant functions again satisfies the strict Mittag-Leffler condition so that ${\lim\limits_{\leftarrow}}^1H^0(M_n)=0$. This means that $H^1(M)=\lim\limits_{\leftarrow} H^1(M_n)$ is always true. Another consequence is $H^k(M)=\lim\limits_{\leftarrow} H^k(M_n)$ if all cohomologies $H^{k-1}(M_n)$ are finite dimensional (because this implies a Mittag-Leffler condition).

However, in general the question remains open (and I would rather expect a counterexample).

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    $\begingroup$ For the record, a proof of Mittag-Leffler for $H^0$. Let $M$ be relatively compact in $N$, with $N$ a manifold. I claim that the image of $H^0(N) \to H^0(M)$ is finite dimensional; this clearly implies ML. Let $N_i$ be the connected components of $N$. I claim that only finitely many of the $N_i$ meet $M$. Suppose, instead, that $x_i$ is an infinite sequence in $M$ with each $x_i$ in a different $N_i$. Let $y \in N$ be a limit point of the $x_i$ with $y \in N_0$. But then $N_0$ contains a ball a round $y$, so $x_n \in N_0$ for $n$ sufficiently large. $\endgroup$ Aug 15, 2017 at 21:10

1 Answer 1


$\def\RR{\mathbb{R}}\def\Hom{\mathrm{Hom}}$This seems too simple, so please tell me what I'm missing. We first prove the corresponding result in homology. Let $C_i(X)$ be the group of singular $i$-chains in $X$. Then $C_i(M) = \lim_{j \to \infty} C_i(M_j)$ (because $i$-simplices are compact, so the image of any finite number of them must lie in some $M_j$). Homology commutes with direct limits, so we deduce that $H_i(M) = \lim_{j \to \infty} H_i(M_j)$.

Now, the universal coefficient theorem tells us that $H^i(M, \RR) = \Hom(H_i(M), \RR)$ and the same for $M_j$. So the question is whether $\Hom( \ , \RR)$ turns direct limits into inverse limits, and the answer is yes.

Both the universal coefficient theorem, and the identification of singular and de Rham cohomology, work for general manifolds.

I also have a direct proof of Mittag-Leffler for $H^q(M_0) \leftarrow H^q(M_1) \leftarrow \cdots$.

Lemma Let $M$ be a manifold (PL or better) and let $K$ be a compact subset of $M$. Then $H^q(M) \to H^q(K)$ has finite dimensional image.

Proof Take a PL triangulation of $M$. For each face $\sigma$ of the triangulation, let $U_{\sigma}$ be the union of the relative interiors of those faces $\tau$ containing $\sigma$. So the $U_{\sigma}$ form an open cover of $M$.

As $K$ is compact, it can be covered by finitely many $U_{\sigma}$. Let $N$ be the union of the closures of those $U_{\sigma}$. So $K \subset N$ and $H^q(M) \to H^q(K)$ factors through $H^q(N)$. But $N$ is a finite simplicial complex, so $H^q(N)$ is finite dimensional. $\square$

Remark Actually, the union of the $U_{\sigma}$ also has finite cohomology, without taking the closure, but I didn't see a one line way of saying it.

Now, let $M_0 \subset M_1 \subset M_2 \subset \cdots$ be an ascending chain of manifolds, with the closure $\overline{M_i}$ (in $M_{i+1}$) compact. Then $H^q(M_{i+1}) \to H^q(M_i)$ factors through $H^q(\overline{M}_i)$, so it has finite dimensional image. For each $j \geq i+1$, the image of $H^q(M_j)$ lies in that finite dimensional image. So the Mittag-Leffler condition holds and $\lim_{\infty \leftarrow j}^1 H^q(M_j)=0$.

  • $\begingroup$ I think you could also do this in a different way using general nonsense. The suspension spectrum functor is a left adjoint, and so preserves colimits. $\pi_0 \operatorname{Maps}( - , H\mathbb{R})$ turns homotopy colimits into limits. One then reduces to showing that $M$ is the homotopy colimit of the $M_n$'s but this is actually also a general nonsense result. See mathoverflow.net/questions/187866/… $\endgroup$
    – Joe Berner
    Aug 16, 2017 at 2:09
  • $\begingroup$ I believe that this is correct (and simple modulo de Rham's theorem) -- thanks a lot. It implies ${\lim\limits_{\leftarrow}}^1H^{k-1}(M_n)=0$ but I still don't see this directly. $\endgroup$ Aug 16, 2017 at 8:01
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    $\begingroup$ @JoeBerner The fact that $M$ is the hocolim of $M_n$'s is true but it is a nontrivial result. It subsumes both the Seifert-van Kampen theorem and the excision theorem for singular homology (and in fact it is more or less equivalent to the combination of those two theorems). Calling it a "general nonsense" result seems a bit dismissive. $\endgroup$ Aug 16, 2017 at 16:36

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