TITLE: fundamental group of $S^1$ is isomorphic to fundamental group of $\mathbb{C}-\{0\}$
QUESTION [0 upvotes]: How to show fundamental group of $S^1$ is isomorphic to fundamental group of $\mathbb{C}-\{0\}$ ?
PS: I'm very beginner in algebraic topo. I saw that both of above groups are isomorphic to $\mathbb{Z}$ that's why I want to know how set up a direct isomorphism between $\pi_1(S^1)$ and $\pi_1(\mathbb{C}-\{0\})$ without using $\mathbb{Z}$

REPLY [3 votes]: Homotopy equivalences induce isomorphisms on the fundamental group. In particular, if one space $X$ is a (strong) deformation retract of another space $Y$, we have $\pi_1(X)\cong \pi_1(Y)$. There is a natural deformation retraction $F$ from $\Bbb C - \{0\}$ to the unit circle where $$F(t,z)= t(\frac z{|z|})+(1-t)z $$
This means that the inclusion of the unit circle into $\Bbb C-\{0\}$ is a homotopy equivalence and induces an isomorphism on $\pi_1$.
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