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 $b1$, where $b$ is the number of boundary components. is this true? (i suspect that $X$ is homootopy equivalent to a bouqet of cicles.

Hey Peter, welcome to math.se. I edited your question since 'algebraic groups' means something different than the content of your question (they are groups in algebraic geometry). I've edited your tags accordingly. Best of luck! – Alex Youcis Oct 10 '15 at 21:30

What does "boundary components" mean for an open subset in the plane? Your suspicion is correct: this follows from the uniformization theorem and the fact that $K(\pi,1)$s are unique up to (weak) homotopy equivalence. – Oct 10 '15 at 22:11

@AlexYoucis oups, yeah I wanted the tag "algebraic topology". I must have been a misclick. Mike what I mean is the number of connected components of $\partial X\subset\mathbb{R}^2$. I have come to think of this because it is known that for (bounded) $X$ we have $X$ simplyconnected iff $b=1$. – Peter Oct 10 '15 at 22:45

2The fundamental group of any open subset of the plane is countable; take $X$ to be the complement of the cantor set for a counterexample to your conjecture. If the rank is finite, I think I can prove you're correct. – Oct 10 '15 at 22:51

@MikeMiller Strangely I cannot find this statement elsewhere on the internet. But I think it is very plausible – Peter Oct 10 '15 at 23:52

@Peter: I'll write up my proof tonight or tomorrow for you. – Oct 10 '15 at 23:54

@MikeMiller: Thanks, i look forward to it. – Peter Oct 11 '15 at 10:48
2 Answers
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 $R^2X$ 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.
 83,864
 5
 97
 192

I don't think you need the complement to be nice. 0th Cech cohomology of a Hausdorff space counts components, so this is all you need. – Oct 12 '15 at 17:32

@MikeMiller: Quasicomponents, see http://math.stackexchange.com/questions/317882/isthereahomologytheorythatcountsconnectedcomponentsofaspace – Moishe Kohan Oct 13 '15 at 17:09

I meant to say compact Hausdorff, where quasicomponents and components agree; see the nLab. In any case if we're doing our Alexander duality in the sphere instead of $\Bbb R^2$ the complement of $X$ is closed hence compact hence we're ok. – Oct 13 '15 at 17:28
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 MayerVietoris 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 $\chi(\Sigma_{0,n}) = 2n$. 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.