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How do we compute the $k$-th rational homology group of the connected sum $\#^{\infty} S^{\infty}$, and hence the sequence of rational Betti numbers ${b_k\left(\#^{\infty} S^{\infty};\mathbb{Q}\right)}=\text{rk} H_k\left(\#^{\infty} S^{\infty};\mathbb{Q}\right)$? Since finding the homology of a sphere is trivial, I am essentially asking how to find the homology of a union of topological spaces?

Context: The space under consideration is embedded in $\mathbb{R}^{\infty}$. In particular, if we consider an infinite-dimensional Hilbert manifold $M$, then we patch it via an atlas $\bigcup_{\alpha\in A}\mathcal{U}_{\alpha}$ for $A$ countable, but infinite and where $\mathcal{U}_{\alpha}$ is $S^{\infty}$,$\forall\alpha\in A$.

Thanks in advance!

Sergio Charles
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    What do you mean by this union? In what ambient space? – Eric Wofsey Aug 04 '18 at 00:15
  • Embedded in $\mathbb{R}^{\infty}$. In particular, if we consider an infinite-dimensional manifold $M$, then we patch it via an atlas $\bigcup_{\alpha\in A}\mathcal{U}{\alpha}$ for $A$ countable, but infinite and where $\mathcal{U}{\alpha}$ is $S^{\infty},\forall\alpha\in A$. @EricWofsey – Sergio Charles Aug 04 '18 at 00:28
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    Every manifold is a union of contractible charts but most have interesting homology. You can say nothing if you don't know pairwise (triple, etc) intersections of your subsets. –  Aug 04 '18 at 00:53
  • @MikeMiller Couldn't we choose an "atlas" such that $\mathcal{U}{\alpha}\cap\mathcal{U}{\beta}=\emptyset$ for all $\alpha,\beta$ and then the homology would simply be $H_k\left(\sqcup_{\alpha\in A}\mathcal{U}{\alpha};\mathbb{Q}\right)=\oplus{\alpha\in A}H_k(\mathcal{U}_{\alpha};\mathbb{Q})$? – Sergio Charles Aug 04 '18 at 01:38
  • If they pairwise intersect trivially that is a disjoint union. –  Aug 04 '18 at 01:55
  • @MikeMiller Please look at my idea that I added above. – Sergio Charles Aug 04 '18 at 02:14
  • The tangent bundle is not a disjoint union of charts. –  Aug 04 '18 at 02:17
  • Sorry, a mental lapse on my part. @MikeMiller – Sergio Charles Aug 04 '18 at 02:19
  • @MikeMiller I meant a connected sum. – Sergio Charles Aug 04 '18 at 07:37
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    For every $k$ the homology of $#^k S^\infty$ is trivial. Note that $S^\infty$ is contractible as well as $S^\infty\setminus{\mathrm{pt}}$. Use Mayer-Vietoris. – Thomas Rot Aug 06 '18 at 12:21

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