Fundamental group of $\mathbb{R}^2 - \{0\}$

Question

I want to prove that the fundamental group of $\mathbb{R}^2 - \{0\}$ is isomorphic to $\mathbb{Z}$.

Given that I know $\pi(S^1) \simeq \mathbb{Z}$, and that if $A \subset X$ is a retract of $X$ implies $j_{*}: \pi_{1}(A, a_0) \to \pi_{1}(X, x_0)$ the induced homomorphism is injective, is the following reasoning correct?

I show $S^1$ is a retract of $\mathbb{R}^2 - \{0\}$ using the map $x \mapsto \frac{x}{||x||}$.

This then implies that $\mathbb{Z}$ is imbedded in $\pi_{1}(\mathbb{R}^2 - \{0\})$.

Letting $\gamma:[0,1] \to \mathbb{R}^2 - \{0\}$ be a loop, we can count the number of times it intersects itself, say $n$, which must be finite. We divide $\gamma$ into $n+1$ loops; each of these is of course homotopic to a loop around $S^1$. Their product then is homotopic to a loop that goes around the circle $n+1$ times, which means that $j_*$ is onto.

Is this alright, and can it be made simpler?

Answer 1

The spaces are homotopy equivalent! Consider the inclusion $\iota : S^1 \to \mathbb R^2\smallsetminus 0$ and the retraction $\pi : \mathbb R^2\smallsetminus 0 \to S^1$ you defined. Certainly $\pi \iota = 1_{S^1}$, and you can check that $\iota\pi$ is homotopic to the identity by means of the homotopy $$H : [0,1] \times\mathbb R^2\smallsetminus 0\to \mathbb R^2\smallsetminus 0$$ $$H(t,x) = tx+(1-t) x/\| x\|$$

This does not pass through $0$ because $1-1/t<0$ if $0<t<1$; geometrically, we are joining $x$ and its projection onto the circle, and this segment does not pass through the origin.

Another way to go around this ---which is what you can do to identify $\pi_1(S^1)$--- is to note that the complex exponential map $\mathbb C\to \mathbb C\smallsetminus 0$ is a covering and exhibits $\mathbb C$ as the universal covering of $\mathbb C\smallsetminus 0$. Because $\exp f(z) = \exp z$ means that $f$ is of the form $z+2\pi i k,k\in \mathbb Z$, this shows that the group of deck transformations of this covering is $\mathbb Z$, whence the result.

Answer 2

@Pedro Tamaroff

Here is my answer to your request. References could be to Topology and Groupoids, to Categories and Groupoids, and a recent paper Modelling and Computing Homotopy Types:I.

The point is that the usual van Kampen theorem for the fundamental group $\pi_1(X,c)$ on a space with base point $c$ generalises to the fundamental groupoid $\pi_1(X,C)$ on a set $C$ of base points; usually we assume $X=U \cup V$ where $U,V$ are open, and $U,V, U\cap V$ are path connected. For the more general case we assume $C$ meets each path component of $U,V, U \cap V$. The proof of the pushout of groupoids is by verifying the universal property, and in this form is an easy generalisation of the proof of Crowell in 1958.

In the case in point, $X= \mathbb R^2 \setminus \{0\} $ while $U,V $ are given by $(x,y) \in X: y \geqslant 0 $ or $ y \leqslant 0$. These are not open, in $X$ but they are closed cofibrations in $X$. Take $C$ to consist of the two points $(\pm 1,0)$. Note that $U,V$ are contractible, as are each component of $U \cap V$.

Let $\mathcal I$ be the groupoid with two objects $\pm 1$ and exactly one arrow $\iota : - 1 \to +1$, and so only one arrow $+1 \to -1$. This groupoid, which seems "trivial", can be thought of, with the inclusions $\{\pm1\} \to \mathcal I$, as a model of the unit interval in the category of groupoids, giving rise to a notion of "homotopy".

In our case we get up to isomorphism a pushout of groupoids of the form

$$\begin{matrix} \{\pm 1\} & \to & \mathcal I \\ \downarrow & & \downarrow \\ \mathcal I & \to & G \end{matrix} $$

What is $G$? We need a bit more groupoid algebra to add to the right of the above square a "retract" square which is also a pushout

$$\begin{matrix} \mathcal I & \to & \{+1\} \\ \downarrow & & \downarrow \\ G & \to& H \end{matrix} $$ giving the composite pushout

$$\begin{matrix} \{\pm 1\} & \to & \{+1\} \\ \downarrow & & \downarrow \\ \mathcal I & \to & H \end{matrix} $$ and from this you show easily that $H \cong \mathbb Z$. This is because the groupoid $\mathcal I$ plays the same role in the category of groupoids as the additive group $\mathbb Z $ plays in the category of groups.

This may seem too much work, but is of course preparing for the wider and potential use of non connected spaces, and so of the algebra of groupoids, in geometric topology. For example, we might have $X$ connected but $U,V$ have $500$ components, and $U \cap V$ has many more!

The argument is also that the use of many base points and the algebra of groupoids is more intuitive, and so easier for students, than being tethered to a single base point. I again refer to the link in my comment.

Further, some of the groupoid methods extend to higher dimensions. See the paper referred to for more references. This is part of the general research theme of "Higher Structures in maths and science" (do a web search).