# Fundamental group of the complement of cell subcomplexes

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.

• 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. – HenrikRüping Nov 11 '19 at 10:56
• 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. – 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).