Given a regular CW complex stucture on a manifold $C$ of dimension $n$ and a subcomplex $D$ of dimension $n-2$, I want to compute the fundamental group of the complement $\pi_1(C\setminus D)$. A procedure is described in section 3.2 of this article: $\pi_1(C\setminus D)$ is generated by the $n-1$ cells in $C$ and the relations are given by the $n-2$ cells in $C\setminus D$. This fact is not proven there and regarded as "well known".

Can you give me a reference? or a proof?

EDIT: Since the formulation of the article is false (see comment below), I added the hypothesis that $C$ is a differentiable manifold.

  • $\begingroup$ Why should this work ? if $D$ is empty, we would just get the fundamental group of $C$ which also depends on the one and two skeleton of $C$. I guess what you want to do it to remove a regular neighborhood around your subcomplex, give the new complement the structure of a CW-complex and use the usual way of getting a presentation for the fundamental group using the 1 and 2 cells of that complex. $\endgroup$ – HenrikRüping Nov 11 '19 at 10:56
  • $\begingroup$ This can't be true without further hypotheses on $C$ or on $D$. For example take $n$ to be very large. Let $C$ be a $n$ dimensional regular complex with at least two zero-cells. Suppose that $D$ contains one but not both of these. Then the remaining two-skeleton in $C - D$ will contribute to the fundamental group. $\endgroup$ – Sam Nead Nov 11 '19 at 14:06

Let $B^k$ be the standard $k$ dimensional ball. An $n$-dimensional $k$-handle is a copy of $B^k \times B^{n - k}$. The boundary of a $k$-handle is

$((\partial B^k) \times B^{n - k}) \cup (B^k \times (\partial B^{n - k}))$

We call the first part of the boundary the attaching region; this will glue "down" to smaller (in $k$) handles. The second half of the boundary is attached to by larger handles.

Suppose that $C$ is a compact connected $n$-manifold without boundary. Suppose that $C$ is equipped with a handle structure where

  1. the attaching maps are very nice,
  2. there is exactly one zero-handle, and
  3. there is exactly one $n$-handle.

Then we can compute the fundamental group of $C$ in two ways. We can use the zero-, one-, and two-handles. Or we can dualise and use the $n$-, $(n-1)$-, and $(n-2)$-handles. As suggested in the original post, after dualising, the $n$-handle gives the base point, the $(n-1)$-handles give generators, and the $(n-2)$-handles give relations. If you remove $D$, made up of $n-2$ and smaller handles, then the dual presentation of the fundamental group still holds (but perhaps with fewer relations).


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