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Related to my previous question here. Let $X$ be a topological space and let $H_c^{\bullet}(X)$ denote its singular cohomology with compact supports (rational coefficients). Let $U$ be an open subset of $X$ and $C$ be its complement (which is then closed). I am trying to prove the existence of a long exact sequence

$$ \cdots \to H^i_c(U) \to H^i_c (X) \to H^i_c (C) \to H^{i+1}_c (U) \to \cdots$$

Can someone help me prove this or provide a reference? I know such a sequence follows if there is a short exact sequence of chain complexes: $$ 0 \to C^{\bullet}_C(U) \to C^{\bullet}_C(X) \to C^{\bullet}_C(C) \to 0.$$

Inclusion could perhaps be the first non-zero map, but what could the other one be? Or is this approach not good. Thanks for any help!

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    $\begingroup$ Are you sure of the direction of the mapping? The embedding $i \colon U \hookrightarrow X$ shoud give a mapping $C^\bullet_C(X) \to C^\bullet_C(U)$, with direction reversed. $\endgroup$ Commented Oct 10, 2013 at 17:17
  • $\begingroup$ @GiorgioMossa I am trying to use this which has the directions in the way I put in my post. $\endgroup$
    – Craig
    Commented Oct 10, 2013 at 17:20
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    $\begingroup$ @GiorgioMossa The directions are correct. For cohomology with compact support a map $f$ induces a map $f^*$ in the opposite direction only if $f$ is proper (this gives a map $H_c(X)\to H_c(C)$, btw). On the other hand, for an open embedding one can construct map $i_*$ in the «homological» direction. $\endgroup$
    – Grigory M
    Commented Oct 13, 2013 at 14:37
  • $\begingroup$ That was my doubt since I known that compact support co-homology lives in the category of CW-complex and proper maps I was wondering how could work the constuction. Thanks @GrigoryM :) $\endgroup$ Commented Oct 13, 2013 at 15:25

3 Answers 3

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$H_c(X)=\tilde H(X^*)$ where $X^*$ is the one-point compactification of $X$ — so theorems about cohomology with compact can be deduced from theorems about ordinary cohomology.

In particular, the long exact sequence for the pair $(X^*,C^*)$ in ordinary cohomology gives the desired exact sequence ($H(X^*,C^*)\cong H_c(U)$ by excision).


Anyway, the map $C_c(X)\to C_c(C)$ is the usual restriction. And the kernel of this restriction is $C_c(X,C)$ which is quasiisomorphic to $C_c(U)$ (by excision: $C(X,(X-K)\cup C)\cong C(U,X-(K\cap U))$; cf. «extension by zero» in the de Rham case).

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  • $\begingroup$ Can you explain in more detail the extension by zero map for singular cohomology? The whole theory of cohomology is much clearer to me for De RHam cohomology, but I need the singular version. $\endgroup$
    – Craig
    Commented Oct 14, 2013 at 9:17
  • $\begingroup$ @Craig Well, I mean that a cochain on $U$ with compact support is more or less the same thing as a cochain on $X$ that is zero on all simplices from $C$; but this identification uses excision isomorphism, so one gets a map not quite from $C_c(U)$ but from something quasiisomorphic — hope updated answer is more clear. $\endgroup$
    – Grigory M
    Commented Oct 14, 2013 at 20:16
  • $\begingroup$ // Note, btw, that to apply excision we need $C$ to be closed (i.e. $U$ to be open). $\endgroup$
    – Grigory M
    Commented Oct 14, 2013 at 20:21
  • $\begingroup$ I am sorry but I can not understand what you have done. Do you mind writing out explicitly how excision gives $H(X^*, C^*) \cong H_c(U)$ ?? Assuming that, I can see how the long exact sequence for that pair gives me the right sequence after the first few terms, but near the start it gives $$0\to H^0_c(U) \to \mathbb{Q}\oplus H^0_c(X) \to \mathbb{Q} \oplus H^0_c(C) \to H^1_c(U)\to \cdots.$$ $\endgroup$
    – Craig
    Commented Oct 15, 2013 at 2:02
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    $\begingroup$ @Katherine What about it? If X is compact X* is X\sqcup point and everything works just fine. $\endgroup$
    – Grigory M
    Commented Feb 10, 2019 at 14:30
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The existence of such an exact sequence can be found in Bredon's book "Sheaf Theory" (version McGraw-Hill 1967, section III.1) for the singular cohomology with compact support where one has to assume that the spaces $X$ and $C$ are HLC which means the following: for each $x \in X$ and neighborhood $U$ of $x$ there is a neighborhood $V \subseteq U$ of $x$ depending on $p$ such that $H_p(V) \to H_p(U)$ is trivial where $H_∗$ is the reduced singular homology.

Another proof can be found in Spanier's book "Algebraic Topology" (version McGraw-Hill 1966, section 6.7), for the Alexander cohomology with compact support, assuming the $X$ is a locally compact Hausdorff space. (In fact, one has to look at the proof of Theorem 15.)

Note that HLC implies that Alexander cohomology coincides with the singular cohomology. Also note that CW-complexes are HLC since they are locally contractible. So the sequence exists for example for complex algebraic varieties and closed subvarieties.

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For what it's worth, when $X$ is a closed compact oriented manifold and $C$ a closed oriented submanifold of codimension $r$, then by Poincare duality, the sequence is isomorphic to the homology sequence

$$\cdots\to H_k(U)\to H_k(X)\to H_{k-r}(C)\to H_{k-1}(U)\to \cdots$$ which is just the long exact sequence of the pair $(X,U)$ by noticing the Thom isomorphism $$H_{k}(X,U)\cong H_{k-r}(C), ~\alpha\mapsto \alpha\cap C.$$

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