Let $X$ be a topological space and $U_i$ open subsets. If $U_i\subset U_{i+1}$ and $\bigcup^{\infty}_{i=1}U_i=X$. How can I prove that if for infinitely many $j$, the $i$-th homology vanishes $H_i(U_j)=0$, then $H_i(X)=0$?
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$\begingroup$ Could you please write your question correctly? What are $X$, $U_i$, $H_i$?? $\endgroup$– abxCommented Nov 6, 2021 at 7:36
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$\begingroup$ Fixed it as you said $\endgroup$– Mary SusyCommented Nov 6, 2021 at 8:29
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1 Answer
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This is true under some mild assumptions using the relationship between unions of spaces and direct limits of homology groups. See Proposition 3.33 of Hatcher's Algebraic Topology.
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1$\begingroup$ I find these "mild" assumptions rather strong — in particular they are not satisfied if $X$ is compact. $\endgroup$– abxCommented Nov 6, 2021 at 10:50
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5$\begingroup$ @abx If $X$ is compact then the question asked is trivial. $\endgroup$ Commented Nov 6, 2021 at 11:23
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5$\begingroup$ @abx Those conditions are always satisfied if the subspaces $X_\alpha$ (in Hatcher's notation) are open. So this actually gives a complete answer to the question asked. $\endgroup$– WojowuCommented Nov 6, 2021 at 13:05
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