Defining a homotopy on an annulus

Question

I have tried to define a homotopy on an annulus to solve problem ii) of the following exercise.

Let A denote the following annulus in the plane A={$(r,\theta) : r \in [1,2], \theta \in [0,2\pi] $} and let h be a homeomorphism of A defined by

$h(r,\theta)=(r,\theta+2\pi(r-1))$

i) Show that h is homotopic to the identity map.

ii) Take the two paths in A

$\alpha(s)=(s+1,0)$

$\beta(s)=(h{\displaystyle \circ } \alpha)(s)$

Show that if h is homotopic to the identity relative to the two boundary circles of A, then the loop $\alpha'.\beta$ where $\alpha'(s)= \alpha(1-s)$ and

$\ \alpha'.\beta(s)=\left\{\begin{array}{ll} \beta(2s), & s\in [0,\frac{1}{2}] \\ \alpha'(2s-1), & s \in [\frac{1}{2},1]\end{array}\right. . $

is homotopic relative to {0,1} to the constant loop at the point (1,0).

iii) Show that $\alpha'.\beta$ represents a non-trivial element of the fundamental group of A.

I tried to solve ii) as follows:

Let e denote the constant loop at the point (1,0). Since h is homotopic to to the identity relative to the two boundary circles of A and homotopy is well-behaved with respect to composition of maps we find a homotopy F' such that

$F''(s,0)=\beta(s)$

$F''(s,1)=Id{\displaystyle \circ }\alpha(s)=\alpha(s)$

$F''(0,t)=\beta(0)=(1,0)$

$F''(1,t)=\beta(1)=(2,2\pi)=(2,0)$

for all s,t $\in [0,1]$.

We also find straight-line homotopies

$G''(s,t)=(1-t)\alpha(s)+t(1,0)$

$G(s,t)=(1-t)\alpha'(s)$+t(1,0)

Hence we can define the homotopy

$\ H'(s,t)=\left\{\begin{array}{ll} F''(s,2t), & t\in [0,\frac{1}{2}] \\ G''(s,2t-1), & t \in [\frac{1}{2},1]\end{array}\right. . $

where $H'(s,0)=\beta(s)$ and $H'(s,1)=e(s)$

Then I considered

$\ H(s,t)=\left\{\begin{array}{ll} H'(2s,t), & s\in [0,\frac{1}{2}] \\ G(2s-1,t), & s \in [\frac{1}{2},1]\end{array}\right. . $

So

$H(s,0)=\alpha'.\beta(s)$

$H(s,1)=(1,0) $

$H(0,t)=(1,0)=H(1,t)$

for all $t \in [0,1]$

The problem is that the Homotopies H' and G do not seem to agree on $[0,\frac{1}{2}] \cap [\frac{1}{2},1]$={$\frac{1}{2}$} which would be crucial for proving that H is well-defined. But I do not see another way to define a map H where

$H(s,0)=\alpha'.\beta(s)$

$H(s,1)=(1,0) $

$H(0,t)=(1,0)=H(1,t)$

for all $t \in [0,1]$

Answer 1

As $h$ is supposed homotopic to the identity $\mathsf{Id}$ relative to the two boundary circles $C_1, C_2$ of $A$, there exists a continuous map $H : A \times [0,1] \to A$ such that $H(x,0) = h(x)$, $H(x, 1) = x$ for all $x \in A$ and $H(x,t) = x$ for all $x \in C_1 \cup C_2$.

Then $H(\alpha^\prime,1-t)$ is an homotopy between $\alpha^\prime$ and $\beta^\prime$ where $\beta^\prime(s) = \beta(1-s)$ as:

• $H(\alpha^\prime,1-0)= H(\alpha^\prime,1)=\alpha^\prime$
• and $H(\alpha^\prime,1-1)= H(\alpha^\prime,0)=h \circ \alpha^\prime = (h \circ \alpha)(1-s) = \beta(1-s)$

Also the initial and terminal points of $H(\alpha^\prime,1-t)$ are fixed for all $t \in [0,1]$.

Based on that, you can prove that $\alpha^\prime . \beta$ is homotopic to $\beta^\prime.\beta$ relative to $(1,0)$. And $\beta^\prime.\beta$ is obviously homotopic to the constant loop at $(1,0)$.