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Let $X$ be a topological space, and $Y,Z$ be subspaces. My question is a bit vague and open-ended: when is it the case that, if $Y$ and $Z$ represent the same (nonzero) cycle in homology, then $Y$ and $Z$ have the same topology?

To be a bit more restrictive. Let $X$ be a smooth manifold. Is it always the case that two submanifolds $Y$ and $Z$ representing the same homology (nonzero) cycle must be homeomorphic? Is there some "natural" hypothesis to put on $X$ to make this true?

Let $X$ be a smooth projective variety over the complex numbers. Is it always the case that two algebraic subvarieties $Y$ and $Z$ representing the same (nonzero) cycle (in Chow, in some cohomology theory) must be homeomorphic?

I have the impression that my questions are a bit naive as formulated, so I would also be happy to hear if there is a good context/theory to formulate a nice question along these lines.

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Consider all the dim $k$ submanifolds of the $n$-sphere for $k<n$ ... –  G.C. Jun 17 '13 at 19:24
    
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). –  Mark Grant Jun 17 '13 at 19:29
    
I am sorry, I forgot to include that the represented cycle is not the zero cycle. Does the question look better now? –  calc Jun 17 '13 at 20:25
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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. –  auniket Jun 17 '13 at 20:51

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The answer to the first pair of questions is that there's no reason that such cycles should be homeomorphic. The question about algebraic subvarieties also has a negative answer, but presumably more restrictive questions might have more positive answers.

Let's stick to the case when the ambient space $X$ is an $n$-manifold, and let $d < n$ be the dimension for your homology class. Then you can simply take $Z$ to be the union of $Y$ with a $d$-dimensional sphere lying in a ball disjoint from $Y$. The number of components has changed so $Y$ and $Z$ are not homeomorphic; on the other hand the sphere is null-homologous so you haven't changed the homology class.

The other question depends on context. If you really mean subvariety, then you presumably allow some singularities; there are many examples where a non-singular curve can degenerate into a singular curve; this doesn't change the homology class. A simple example would be in $CP^2$, where a smooth curve of degree 2 (thus homeomorphic to a 2-sphere) can degenerate into a union of two lines (two 2-spheres meeting at a point). If you intended that your subvarieties be smooth, then this is more subtle and perhaps someone more knowledgeable about Chow rings can answer your question.

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This fails systematically, in $P^n}$, the only cohomology invariant is "degree". All sorts of non-homeomorphic varities can be embedded as projective varieties of the same degree. n= 2 d= 3 allows for elliptic curves and rational normal curve of degree 3. This can be gussied up for anything except degree 1 and (sometimes ?) 2. –  aginensky Jun 17 '13 at 22:21
    
oops- comment added in wrong place. The question fails systematically, not the given answer –  aginensky Jun 17 '13 at 22:22
    
You are right, my question is silly. Sorry about that. –  calc Jun 18 '13 at 19:06

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