$X$ is contained in an annulus containing a circle in the annulus.Show that $\pi_1(X)$ contains a subgroup isomorphic to $\mathbb{Z}$.

Question

In $\mathbb{R^2}$ let $C=\{|x|=2\}$ and $A=\{1<|x|<3\}$, let $X$ be a path connected set such that $C \subset X \subset A$. Show that $\pi_1(X)$ contains a subgroup isomorphic to $\mathbb{Z}$.

Well, it is intuitively obvious but how can one argue rigorously? Can I argue like : $C$ itself can be thought of as a path $\gamma$ in $X$ with $\gamma : [0,1] \to X$ is defined by $\gamma(t)=2e^{2\pi it}$. Since $\gamma$ is not nullhomotopic (My question is : does this fact need proof ? If yes, how should I prove that ?) and hence [$\gamma$] generates an infinite cyclic group which is contained in $\pi_1(X)$... $\blacksquare$

Isn't it too handwavy ? thank you in advance for your helpful comments/answers

Answer 1

The inclusions $C\subset X\subset A$ induce maps on the fundamental groups $$\pi_1(C)\to \pi_1(X)\to \pi_1(A).$$ The composition of the two maps is an isomorphism since $C$ is a deformation retract of $A$. It follows that $\pi_1(C)\to \pi_1(X)$ is injective.

Answer 2

The inclusion $X \hookrightarrow A$ induces a homomorphism on fundamental groups $\pi_1(X) \rightarrow \pi_1(A)$. As you said, pick the element $\gamma$ of $\pi_1(X)$ corresponding to $\{|x| = 2\}$ (i.e., a path traversing this circle once counterclockwise). The image of $\gamma$ in $\pi_1(A)$ represents a generator of $\pi_1(A) \cong \Bbb Z$ (why?), so that $\gamma$ must have infinite order in $\pi_1(X)$ (why?)

(As mentioned in the comments above, this all comes down to having previously computed $\pi_1(S^1)$!)

Answer 3

Consider the inclusion maps $C\stackrel{i_1}{\to} X\stackrel{i_2}{\to} A$ which induces a sequence of group homomorphisms $$\pi_1 (C)\stackrel{{i_1}_*}{\to} \pi_1 (X)\stackrel{{i_2}_*}{\to} \pi_1 (A)$$ Note that ${i_2}_*\circ {i_1}_*\colon \pi_1 (C)\to\pi_1 (A)$ is an isomorphism, and so in particular a generator $\alpha$ of $\pi_1(C)$ is not mapped to zero in $\pi_1 (X)$ by ${i_1}_*$. It follows that $\beta={i_1}_*(\alpha)$ is non-zero and so $\langle\beta\rangle$ is a cyclic subgroup of $\pi_1 (X)$. We know $\langle\beta\rangle$ can not be finite because $\beta$ is mapped to a generator of $\pi_1(A)$ under ${i_2}_*$ and so can't have torsion. This then means $\langle\beta\rangle$ must be infinite cyclic and so $\pi_1 (X)$ contains a subgroup isomorphic to $\mathbb{Z}$.

Answer 4

The fact that it is contained in an annulus means in particular that it doesn't contain the origin. Thus there is a well-defined retract $r:X \to C$ given by

$$r(x) = 2\frac{x}{\|x\|}$$

If you haven't seen it before, the notion of retract is that the inclusion $i:C \to X$ and the surjection (retract) $r:X \to C$ are partial inverses in the sense that $r \circ i = \text{id}_C$.

It is then a general fact that if we have $Y \subset Z$, with $r:Z \to Y$ a retract, then the induced map $i_*:\pi_1(Y) \to \pi_1(Z)$ is an injection. as $r_*$ is a left inverse.