A colleague asked me a topology question which comes down to this: Suppose that $M$ is a smooth $n$manifold, and $C\subset M$ is a closed set such that $H_{np}(MC)\to H_{np}(M)$ is not surjective. What can you say about $C$? Clearly in some sense it is at least $p$dimensional, but in what sense? For example, must it have a subset homeomorphic to $\mathbb R^p$?

$\begingroup$ That's what the topological duality theorems are about. $\endgroup$ – Włodzimierz Holsztyński Apr 18 '14 at 0:55

$\begingroup$ What are the topological duality theorems? $\endgroup$ – Tom Goodwillie Apr 18 '14 at 1:48

$\begingroup$ First there was the absolute (just for the spaces, not for subsets) Poincare duality theorem. Then there was AlexanderPontryagin theorem for subsets of a manifold. On the later occasion Pontryagin introduced his duality theorem for topological groupsinitially it was about compact abelian groups versus discrete abelian groups. (This was generalized to locallyt compact abelian groups Egbert van Kampen in 1935 and André Weil in 1940see wikipedia). An early result about dissecting $\mathbb R^n$ by a compact subset was obtained by Karol Borsuk. Etc. (you need to ask not me but a specialist). $\endgroup$ – Włodzimierz Holsztyński Apr 18 '14 at 4:26

$\begingroup$ Certainly some kind of $p$th Cech cohomology of $C$ is nontrivial. My question is, what does this imply geometrically about $C$? $\endgroup$ – Tom Goodwillie Apr 18 '14 at 12:23
For sure you can not expect a subset homeomorphic to $\mathbb R^p$.
Say, take $p=1$ and $M=\mathbb T^2$. Note that there is an open embedding of cylinder $f\colon (0,1)\times \mathbb S^1\to\mathbb T^2$ such that the complement $\Sigma=\mathbb T^2 \backslash \mathrm{Im}f$ is a pseudocircle. In particular $\Sigma$ does not contain a subset homeomorphic to $\mathbb R$. Clearly $$\mathbb Z=H_{1}(\mathbb T^2\Sigma)\to H_{1}(\mathbb T^2)=\mathbb Z^2$$ is not surjective.
P.S. The pseudocircle can be constructed as an intersection of nested crooked chains of dics. Here is the image of a chain crooked in a circular chain with 6 links from here; it gives the second iteration in the construction.

$\begingroup$ Why is the complement homeomorphic to a cylinder? $\endgroup$ – Igor Belegradek Apr 18 '14 at 14:54

1$\begingroup$ Thank you, Anton. I did not know that there were such strange examples in such low dimensions. $\endgroup$ – Tom Goodwillie Apr 18 '14 at 16:25

$\begingroup$ Anton, what do you mean by statement: For sure you can not expect a subset homeomorphic to $\mathbb R^p$ ? $\endgroup$ – Włodzimierz Holsztyński Apr 18 '14 at 18:18

1$\begingroup$ @WlodzimierzHolsztynski: search the web on "pseudocircle" and "continuum". $\endgroup$ – Igor Belegradek Apr 18 '14 at 19:43

1$\begingroup$ I gather, Cech cohomology of the pseudocircle are the same as for the circle, so by Alexander duality its complement in the plane has two components and has first homology isomorphic to $\mathbb Z$. By the classification of surfaces this means that the complement's components are open disk and open annulus. Joining them by a handle gives the desired embedding into the $2$torus. The only thing I am not sure about is whether the first Cech cohomology is $\mathbb Z$, but if this weren't true, why would it be called pseudocircle? $\endgroup$ – Igor Belegradek Apr 18 '14 at 21:02