I've started learning about the degree of a map in my algebraic topology course. Looking at text references it seems as though my professor is giving a less general and more concrete description of the degree of maps from $S^1\rightarrow S^1$ but I am having trouble with the computations. Let me write out the set up.
Definition:Let $p:\mathbb{R}\rightarrow S^1$ be the exponential mapping $p(x)=(\cos(2\pi x),\sin(2\pi x))$.
Theorem (The Path Lifting Lemma). If $\alpha:[0,1]\rightarrow S^1$ is a continuous function and $x\in\mathbb{R}$ is such that $p(x)=\alpha(0)$, then there exists a unique continuos function $\tilde{\alpha}:[0,1]\rightarrow\mathbb{R}$ called a lift of $\alpha$ such that $p(\tilde{\alpha}(s))=\alpha(s)$ for all $s\in[0,1]$ and $\tilde{\alpha}(0)=x$.
Definition: Let $\pi:[0,1]\rightarrow S^1$ be the quotient map obtained by restriction the map $p$ to the interval $[0,1]$.
Definition: Given a continuous map $f:S^1\rightarrow S^1$, let $\alpha=f\circ \pi$ and $\tilde{\alpha}:[0,1]\rightarrow\mathbb{R}$ be a lift. We call the degree of $f$ denoted $deg(f)$ the integer $\tilde{\alpha}(1)-\tilde{\alpha}(0)$.
Now the particular exercise I am stuck on is computing the degree of the maps $f,g:S^1\rightarrow S^1$; $f(x,y)=(-x,-y)$ and $g(x,y)=(x,-y)$
For $f$ I have the following computations. It is simple enough to compute $\alpha(0)=(f(\pi(0))=(-1,0)$. Then I can let $x=1/2$ so that $p(1/2)=(-1,0)=\alpha(0)$. This also makes $\tilde{\alpha}(0)=1/2$ as stated in the theorem. What I am stuck on is what is the correct way to compute $\tilde{\alpha}(1)$. I know that $p(\tilde{\alpha}(1))=\alpha(1)=(-1,0)=\alpha(0)$. This means that $\tilde{\alpha}(1)$ needs to be any odd integer divided by 2 since $p(k)=(-1,0)$ only when $k$ is an odd integer divided by 2 but it seems to me that any odd integer will work. If anybody can point out what I am missing or what my mistake is it will be greatly appreciated.
Best Answer
Let us write $q : [0,1] \to S^1$ for the quotient map in order to not confuse it with the real number $\pi$. We have $q = p \mid_{[0,1]}$.
Identifying $(x,y) \in \mathbb R^2$ with $x + iy \in \mathbb C$, we can write $p(x) = e^{2\pi ix}$. Then $f(z) = -z$ and $g(z) = \overline z$.
For $\alpha = f \circ q$ we have $\alpha(t) = -e^{2\pi i t}$. A lift is given by $\tilde \alpha(t) = t + 1/2$ because $p(\tilde \alpha(t)) = e^{2\pi i(t + 1/2)} = e^{2\pi it} e^{\pi i} = - e^{2\pi it} = \alpha(t)$. Thus $\deg(f) = \tilde \alpha(1) - \tilde \alpha(0) = 1$.
For $\alpha = g \circ q$ we have $\alpha(t) = \overline{e^{2\pi i t}} = e^{\overline{2\pi i t}} = e^{-2\pi i t}$. Obviously a lift is given by $\tilde \alpha(t) = -t$. Thus $\deg(g) = \tilde \alpha(1) - \tilde \alpha(0) = -1$.