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Jun 18, 2013 at 19:05 vote accept calc
Jun 18, 2013 at 19:05 vote accept calc
Jun 18, 2013 at 19:05
Jun 17, 2013 at 20:51 comment added pinaki Ehresmann's Theorem and theorems of Varchenko (in the algebraic case) and Verdier in analytic (and possibly more general categories) case give a somewhat positive statement: if you take a "family" of homologous cycles, then "almost all" of them will be homeomorphic.
Jun 17, 2013 at 20:25 comment added calc I am sorry, I forgot to include that the represented cycle is not the zero cycle. Does the question look better now?
Jun 17, 2013 at 20:23 history edited calc CC BY-SA 3.0
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Jun 17, 2013 at 19:31 answer added Danny Ruberman timeline score: 1
Jun 17, 2013 at 19:29 comment added Mark Grant The answer to the question in your second paragraph is 'no'. For any $X$ of dimension at least $n$, one can always find non-homeomorphic submanifolds $Y$ and $Z$ of dimension $n-1$ which represent the zero element in $H_{n-1}(X)$. Perhaps bordism theory offers a context in which to formulate such questions (and give negative answers, I fear).
Jun 17, 2013 at 19:24 comment added G.C. Consider all the dim $k$ submanifolds of the $n$-sphere for $k<n$ ...
Jun 17, 2013 at 18:51 history asked calc CC BY-SA 3.0