# Fundamental group of $\mathbb{R}^n\backslash \{0\}$

I am wondering about what the fundamental group of $\mathbb{R}^n \backslash \{0\}$ or more generally $\mathbb{R}^n \backslash U$ where $U$ is a subset of $\mathbb{R}^n$ for $n>1$. For $n=1$ I understand that the fundamental group is not trivial since the resulting space is not simply-connected. In all other case though the space is simply connected. Thus, logic says that the fundamental group is trivial. Therefore how do we classify spaces that we subtract various points out of them. E.g. how we can tell the difference between $\mathbb{R}^n \backslash U$ and $\mathbb{R}^n \backslash V$ for different sets $U,V$ (with different number of elements as well)? Is this the reason we need homology?

For $n = 1$, the space $\mathbb{R} \setminus \{0\}$ is the disjoint union of two intervals, so both connected components are contractible and have trivial fundamental group.

When $n = 2$, you can see that $\mathbb{R}^2 \setminus \{0\}$ deformation retracts onto $$S^1 = \{ (x,y) \in \mathbb{R}^2 : x^2 + y^2 = 1 \},$$ the retraction being $z \mapsto \frac{z}{\|z\|}$. This is the standard circle, in particular its fundamental group is $\mathbb{Z}$ as usual. This space is not simply connected as you claimed.

More generally, the space $\mathbb{R}^n \setminus \{0\}$ deformation retracts onto $$S^{n-1} = \left\{ (x_1, \dots, x_n) \in \mathbb{R}^n : \sum x_i^2 = 1 \right\}.$$ This is the standard $n$-sphere, and it's a classical exercise that this space is simply connected when $n > 2$.

Now when $U \subset \mathbb{R}^n$ is an arbitrary set, even open, you cannot say anything at all about the fundamental group of $\mathbb{R}^n \setminus U$, and not even its homology, without knowing anything about $U$. Indeed it's a theorem of Whitehead that any closed manifold embeds into $\mathbb{R}^n$ for some $n$; if you take $U$ to be the complement of the image of the embedding, you can see that $\mathbb{R}^n \setminus U$ can have pretty much any fundamental group or homology you can imagine (there are some restrictions for closed manifolds of course, but you can imagine even more exotic spaces).

Compared to simply the fundamental group, homology also allows you to detect higher dimensional structure of your space: $\pi_1$ doesn't distinguish the $n$-spheres $S^n$ for $n \ge 2$, but these spaces are intrinsically very different. But the higher homotopy groups $\pi_n$ also detect higher dimensional structure, and do distinguish the $n$-spheres.

From my perspective, the main reason you want to use homology instead of (or in addition to) homotopy groups is because of the Mayer–Vietoris exact sequence, an analog of the Seifert–van Kampen theorem for homology. This allows you to break up your space in smaller pieces. See this question and its excellent answer by Ronnie Brown.

• I am confused. Imagine I start a path at $x$ far away from $0$ at $\mathbb{R}^n$. Then locally the fundamental group at that point is trivial. But on the other hand, you re right in the sense that if I try to take a loop near $0$ indeed I get that the fundamental group is $\mathbb{Z}$. Also, how do you classify such spaces in principle (imagine you know what these sets $U,V$ are)? – Marion Jan 23 '15 at 17:48
• @Marion Well, it's possible that $\mathbb{R}^n \setminus U$ is bounded for example, take $U = \{ \|x\| > 1 \}$ for example. Then you can't start "very far away" from $0$. Another thing to consider is that even if locally you can trivialize the fundamental group, globally there might still be nontrivial loops. Think about a solid torus (a "donut"): locally it's contractible, but in globality there's a loop that cannot be trivialized. As for classification, as I explained it's at least as hard as classifying all manifolds: something that's simply impossible in high dimensions. – Najib Idrissi Jan 23 '15 at 17:51
• I see what you mean "globally" and I will agree with you for $\mathbb{R}^2$. I "see" it but, surly for $\mathbb{R}^n$ with $n \geq 3$ the fundamental group is trivial. Simple-connectedness requires/implies that the fundamental group is trivial. Correct me if mistaken. Also thanks for your wonderful answer. – Marion Jan 23 '15 at 17:56
• @Marion Yes, a space is simply connected iff it is path-connected and its fundamental group is trivial. It is true that $\mathbb{R}^n \setminus \{0\}$ is simply connected for $n \ge 3$, but not for a general set $U$! For example, take the complement of a circle embedded in $\mathbb{R}^3$... – Najib Idrissi Jan 23 '15 at 17:57
• @Marion It's almost that. Homology uses singular simplices to "detect" higher dimension holes. It produces powerful invariants, but unfortunately that's not enough to completely classify spaces. For example there exist spaces $X$ such that $H_k(X) = H_k(S^n)$ for all $k$, but $X$ and $S^n$ are not homotopy equivalent: the so-called homology spheres. If you then consider $X \setminus \{x\}$, that space has $H_k = 0$ for all $k$ but is not contractible! Homology is still an extremely useful tool, though. – Najib Idrissi Jan 23 '15 at 18:07

You are right, in higher dimensions ($n>2$) the (first) fundamental group of $\mathbb{R}^n$ minus a point is trivial. This is, however, not true for the removal of an arbitrary subset. Think of removing a tubular shape or just a circle from $\mathbb{R}^3$.

Other means of classification are so called higher homotopy groups, where you look at homotopy classes of maps from spheres of any dimension into the space you want to classify. See (almost) any book on algebraic topology.