Fundamental group of $\{z\in\mathbb{C} \mid |z|\geq 1\}$

Question

I was wondering how I can prove that $A:=\{z\in\mathbb{C} \mid |z|\geq 1\}$ is not homeomorphic to $\mathbb{C}^*$ using fundamental groups.

I do not very much about algebraic topology, I can prove that $\mathbb{C}^*$ is homotopic (equivalent) to $S^1$ so that $\pi_1(\mathbb{C}^*)$ is isomorphic to $\mathbb{Z}.$

Not sure about $A.$ What is the fundamental group of $A$?

I can prove that $A$ is not homeomorphic to $\mathbb{C}^*$ using one point compactification. I am just curious.

Answer 1

---

Note that $A$ is strong deformation retract into $S^1$ via $D:A×I\rightarrow S^1,D(a,t)=(1-t)a+t\frac{a}{||a||}$ . Therefore fundamental group of $A$ isomorphic to fundamental group of $S^1$.

---

For more information, now I prove here that $A:=\{z\in C : |z|≥1\}$ is not homeomorphic to $C^*:=C-\{0\}$ using fundamental group.

If possible let $C^*$ is homeomorphic to $A$ via some homeomorphism $f:A\rightarrow C^*$ then $B:=A-\{1\}$ is homeomorphic to $D:=C-\{0,a\}$ for some $a\in C^*$. Note that B is deformation retract to subspace $ E:=\{z\in C : ||z||= 2\}$ via $g:B×I:\rightarrow E, g(b,t)=(1-t)b+2t\frac{b}{||b||}$ i.e fundamental group of $B$ again isomorphic to $Z$. But notice that $D$ is deformation retract to a subspace $D'\ homeomorphic\ to\ S^1\lor S^1$ and fundamental group of $S^1\lor S^1$ is $Z*Z=a\ free\ group$ which is certainly non-abelian.

Answer 2

It is a deformation of $S^1$ (consider $h_t(z)=tz+(1-t){z\over{|z|}}$) so its fundamental group is $\mathbb{Z}$.