Given a pair of closed paths $\alpha$ and $\beta$ . $\alpha(t)=2ie^{it/2}$ for $0\leq t\leq4\pi$, and $\beta=\beta_1+\beta_2$ with $\beta_1(t)=-1+e^{it}$ and $\beta_2(t)= 1-e^{it}$ for $0\leq t\leq2\pi$,
(1) $\gamma=\beta_2+\beta_1-\beta_2-\beta_1, A=\Bbb C \setminus \{-1, 1\}$ . How to show that $\gamma$ is not contractible in $A$?
(2) $B=\Bbb C \setminus \{-1/2, 1/2\}$. Decide whther $\alpha$ and $\beta$ are freely homotopic in $B$?
A free homotopy between two closed paths $ \lambda,\mu:[a,b]⟶U $ is a continuos function $H:[a,b]×[0,1]⟶U$, such that $H(t,0)=\lambda(t), H(t,1)=\mu(t) $ and $H(a,s)=H(b,s),∀t∈[a,b],∀s∈[0,1]$.
$\gamma$ is contractible in $A$ means $\gamma$ is freely homotopic in $A$ to a constant path.
$\alpha$ and $\beta$ are homologous in $U$ if $\alpha$ and $\beta$ have same winding numbers around every point in $\Bbb C\setminus U$
Source: An introduction to complex function theory, Bruce P. Palka P204, P213
It is obvious that $\alpha$ and $\beta$ are homologous in $B$($\alpha$ and $\beta$ have same winding numbers around $-1/2, 1/2$), I think $\alpha$ and $\beta$ are not freely homotopic in $A$. But how to prove this?
Thanks in advance