TITLE: Dimension of a set detected by a homology class
QUESTION [7 upvotes]: 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 $H_{n-p}(M-C)\to H_{n-p}(M)$ 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 $\mathbb R^p$?

REPLY [5 votes]: 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 pseudo-circle.
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 pseudo-circle 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.