Possible groups for fundamental group of $\mathbb{R}$

Question

I want to know if there exist a topology on $\mathbb{R}$ such that $\pi_1(\mathbb{R})\ne (0)$?

If not, we conclude that the fundamental group is a property of the underground set, not the topology it carries, right?

If yes, may I know what topology is it?

Thanks in advance.

Answer 1

If $X$ is any topological space with cardinality $c$ (the cardinality of $\Bbb R$) then there is a topology on $\Bbb R$ making $\Bbb R$ homeomorphic to $X$. Choose $X$ so that $\pi_1(X)$ is non-trivial (for example the plane minus a point).

Answer 2

Yes, there is such a topology. Take a bijection $f$ from $\mathbb R$ onto $S^1$ and define this topology on $\mathbb R$: a subset $A$ of $\mathbb R$ is open if and only if $f(A)$ is an open subset of $S^1$. Then $\pi_1(\mathbb{R})\simeq(\mathbb{Z},+)$.

Answer 3

In fact, for any given finitely presentable group $G$ there is a topology on $\mathbb{R}$ such that $\pi(\mathbb{R})=G$. This follows from two facts:

1. There exists a closed 4-manifold $M$ with a fundamental group isomorphic to any given finitely presentable group $G$.

2. There exists a bijection between $M$ and $\mathbb{R}$, because their cardinalities are the same.