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I am trying to understand the definition of (singular) cohomology with compact supports.

My understanding of singular cohomology goes like this. Let $X$ be a topological space. Define the singular chain group $C_n(X)$ to be the free abelian group generated by singular n-simplicies, which are functions $\sigma : \Delta^n \to X.$ The homology of the complex $C_{\bullet}(X)$ is singular homology. If we define $C^n(X):= \operatorname{Hom}(C_n(X), \mathbb{Z})$ to be the dual of the singular cochain group we get singular co-chains. If we now take the cohomology of this new complex, we get singular cohomology.

Now I am trying to understand cohomology with compact supports, which this source begins to define (on page 7 of the pdf) as such:

Given a (singular, simplicial, cellular) cochain complex $C^{\bullet}$ on a space $X$ , consider the subcomplex $C^{\bullet}_c$ of cochains which are compactly supported: each cochain is zero outside some compact subset of $X$ .

What does this last statement mean? Isn't a cochain a map from $C_n(X)$ to $\mathbb{Z}$? How can take have a support inside $X?$ Please help me clear my deep misunderstanding. Thank you.

Craig
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1 Answers1

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The idea is that a cochain $\varphi \in C^n(X)$ is compactly supported if there's a $K \subseteq X$ compact subset of $X$ such that $\varphi|_{C_n(X \setminus K)} = 0$.

Edit a little remark: for every $K$ compact subset of $X$ there's an embedding $i \colon X \setminus K \hookrightarrow X$ which give rise to a injective embedding of chain complexes $i_* \colon C_\bullet(X \setminus K) \to C_\bullet (X)$, so we can think of $C_n(X \setminus K)$ as being a submodule of $C_n(X)$ and to be exact what I meant above by $\varphi|_{C_(X \setminus K)}$ should be written more formally as $\varphi|_{i_*(C_n(X \setminus K))}$.

So compactly supported co-chain of $X$ are those co-chains in $C^\bullet(X)$ that vanish on all the simplexes that have image contained in a subspace $X \setminus K$ (for some $K$ compact subset of $X$), i.e. those simplexes $\sigma \colon \Delta^n \to X$ that factors through the inclusion map $i \colon X \setminus K \to X$.

You can find out more about this in Hatcher's book Algebraic Topology.

Giorgio Mossa
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  • Let me see if I understand. $C_n(X)$ is the free abelian group generated by functions $\sigma : \Delta^n \to X.$ If we strict ourselves to only the simplicies that are non-zero inside $X\setminus K$, we can consider them as functions $\sigma:\Delta^n \to X\setminus K.$ The free abelian group generated by only these simplicies is denoted $C_n(X\setminus K).$ A cochain is a map from $C_n(X)$ to $\mathbb{Z}$ and it is compactly supported if its restriction to $C_n(X\setminus K) \subset C_n(X)$ is zero. Is that correct & should the superscript in $C^n(X\setminus K)$ actually be a subscript? – Craig Oct 10 '13 at 10:28
  • Yes is correct. Just one remark clearly the requirement is that it must be a compact $K$ such that the cochain vanishes outside that compact, and this compact can be different for any compactly supported cochain. – Giorgio Mossa Oct 10 '13 at 10:33
  • Ahh ok. Thank you very much Giorgio. – Craig Oct 10 '13 at 10:37
  • Sorry, I still don't understand. When I said above "If we strict ourselves to only the simplicies that are non-zero inside $X\setminus K$" , that doesn't quite make sense. $X$ is the image, not the domain. If I change that to "only the simplicies whose image is contained inside $X\setminus K$" then is that the correct definition? – Craig Oct 10 '13 at 11:08
  • Yes, that's correct: I've just noted that there was a little mistake anywayI'm gonna add some details. So stay tuned. – Giorgio Mossa Oct 10 '13 at 14:42
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    I thought that the compact set $K$ could be different for each simplex? The new definition does not seem to allow that. – Craig Oct 10 '13 at 14:59
  • No it doesn't depend on the choice of the compact, it must just exists a compact subset of $X$ out which vanishes in the sense said above. :) Anyway I've edited the answer again. – Giorgio Mossa Oct 10 '13 at 15:02
  • Thank you for your patience, I think I've finally understood. If you have any pointers for this that would be awesome but I'll understand if I've tired you out! – Craig Oct 10 '13 at 15:09
  • @Craig here's Hatcher book, I very good book to start learning some algebraic topology http://www.math.cornell.edu/~hatcher/AT/ATpage.html on page 251 (if I'm not mistaken) you can find a part about co-homology with compact support. – Giorgio Mossa Oct 10 '13 at 15:27
  • Thanks for the link. My previous comment forgot the proper link to the other question: math.stackexchange.com/questions/521429 – Craig Oct 10 '13 at 15:32