Let $h: S^1 \to S^1$ be a continuous map such that $h(x_0)=x_1$. This induces a map $(h_{x_0})_*: \pi_1(S^1,x_0) \to \pi_1(S^1,x_1) $. Now if we choose any other point, say $y_0 \in S^1$ and $h(y_0)=y_1$, then we get a map $(h_{y_0})_*: \pi_1(S^1,y_0) \to \pi_1(S^1,y_1)$.
Degree is defined as follows:
Let $b_0 = (1,0) \in S^1$ and $\pi_1(S^1,b_0)$ be the fundamental group at the base point $b_0$ and $\gamma$ be its genertor. Let $\alpha$ be a path from $b_0$ to $x_0$ and let $\gamma_{x_0}= \bar{\alpha}*\gamma * \alpha$. Then $\gamma_{x_0}$ is the generator of $\pi_1(S^1,x_0)$. Note that $\gamma_{x_0}$ is independent of the path $\alpha$ since the fundamental group is abelian. Now, the map $(h_{x_0})_*(\gamma_{x_0})= d. \gamma_{x_1}$ where $\gamma_{x_1}$ is defined similarly. The integer $d$ is defined as the degree of the map $h$.
How do I see that the degrees of both the maps are the same ? This might be trivial but I am not seeing it right now ! I am stuck at it ! Any hint/answer would really be helpful !
Can anybody help me this ? I am stuck at this ! Any comment/suggestion/hint/answer is welcome !
Thanks
Let $x_0$ belong to $S^{1}$ with $h(x_0)=x_1$ and $h(b_0)=b_1$. Let $α$ and $β$ be paths from $b_0$ to $x_0$ and $b_0$ to $b_1$ respectively. Hence $β*(hoα)$ is a path from $b_0$ to $x_1$.
Hence by definition $\gamma_{b_0} = \gamma$,
$\gamma_{b_1} =$ $[\bar{\beta}]*\gamma * [\beta]$
$\gamma_{x_0}= [\bar{\alpha}]*\gamma * [\alpha]$,
$\gamma_{x_1} = [\overline{\beta*(ho \alpha)}] *\gamma * [\beta*(ho \alpha)]$.
Now let $h_*(\gamma_{b_0})= [\gamma_{b_1}]^{n}$, (i.e, deg of h with base point $b_0$ is $n$.).
$h_*(\gamma_{x_0}) = h_*([\bar{\alpha}]*\gamma * [\alpha])$ $=[ho \bar{\alpha}]*h_*(\gamma_{b_0}) * [ho \alpha]$. $=[\overline{ho \alpha}]*[\bar{\beta}]* \gamma^{n}*[\beta] * [ho \alpha]$ $=\big([\overline{\beta*(ho\alpha)}]*\gamma^{n}*[\beta*(ho\alpha)]\bigr)$ $=[\gamma_{x_1}]^{n}$.
This shows that degree of $h$ is independent of base points.