Rank of fundamental groups of open subsets.

Question

Let $X$ be an open connected subset of the real plane. Then it is known that $\pi_1(X)$ is a free group. Is there a useful formula for the rank of $\pi_1(X)$? I suspect that the rank should be $b-1$, where $b$ is the number of boundary components. is this true? (i suspect that $X$ is homootopy equivalent to a bouqet of cicles.

Answer 1

Studiosus's answer explains how to get the result you're looking for. This post is orthogonal: it's about how to understand the homeomorphism type of connected open subsets of the plane.

Every noncompact manifold without boundary has an exhaustion by compact submanifolds: $M = \bigcup M_i$, where $M_1 \subset M_{2} \subset \dots$ are compact manifolds contained in the interior of the next. (You can prove this by first demonstrating the existence of a proper Morse function on $M$, but it might be best to just take it for granted.)

So let $M$ be your open subset of $\Bbb R^2$. What can $M_0$ be? By the classification of compact surfaces, and the fact that no surface of positive genus (with boundary or not) can embed in $\Bbb R^2$, it has to be $S^2$ with a few open discs deleted. The same is true for all of the $M_{i+1} \setminus \text{int } M_i$. First, a quick calculation: by the Mayer-Vietoris sequence you can see that $\chi(M_{i+1}) = \chi(M_i) + \chi(M_{i+1} \setminus \text{int }M_i).$ (This is essentially because $\chi(S^1) = 0$.)

Let $\Sigma_{0,n}$ be $S^2$ minus $n$ open discs. Then $\chi(\Sigma_{0,n}) = 2-n$. The only $\Sigma_{0,n}$ with positive Euler characteristic and with boundary are the disc $\Sigma_{0,1}$ and the annulus $\Sigma_{0,2}$. Gluing on a disc reduces the number of boundary components by one, and gluing on an annulus doesn't change the homeomorphism type of your manifold. Thus, putting together everything we've said above, because $\chi(M) = 1 - \text{rank}(\pi_1)$ if $\text{rank}(\pi_1)$ is finite, $H_k(M) = \lim_{i \to \infty} H_k(M_i)$, and our assumption that $\text{rank}(\pi_1)$ was finite, we see that the homeomorphism type of an open connected subset of the plane with $\pi_1(M)$ of finite rank is a sphere punctured $\text{rank } \pi_1(M)+1$ times. That is, it's homeomorphic to $\Bbb R^2$ with $\text{rank } \pi_1(M)$ points deleted.

Answer 2

There is a couple of related questions: $H_2$ and $\pi_1$ of open subsets of $\mathbb{R}^2$ and fundamental groups of open subsets of the plane concerning the fact that the fundamental group of each open connected planar set is free of countable (possibly finite) rank, since $R^2-X$ is homotopy equivalent to a connected locally finite graph. If you want to compute rank of the fundamental group, the easiest thing to do is to use homology. Rank of $H_1(X)$ equals the rank of $\pi_1(X)$. On the other hand, by the Alexander duality, $H_1(X)\cong \check{\tilde{H}}^0(R^2 -X)$ (the reduced Chech cohomology). In the case when the complement to $X$ is reasonably nice (say, a finite CW complex), then indeed, the above formula shows that the rank of $\pi_1(X)$ is the number of complementary components minus 1.