Show that if $h,k: S^1\rightarrow S^1$ are homotopic, they have the same degree.

Question

We define the degree of a continuous map $h: S^1 \rightarrow S^1$ as follows:

Let $b_0$ be the point $(1,0)$ of $S^1$; choose a generator $\gamma$ for the infinite cyclic group $\pi_1(S^1,b_0)$. If $x_0$ is any point of $S^1$, choose a path $\alpha$ in $S^1$ from $b_o$ to $x_o$, and define $$\gamma_{x_0} = \hat\alpha(\gamma) = [\overline\alpha] * \gamma * [\alpha].$$ Then $\gamma_{x_0}$ generates $\pi_1(S^1,x_0)$. The element $\gamma_{x_0}$ is independent of the choice of the path $\alpha$, since the fundamental group of $S^1$ is abelian.

Now given $h: S^1\rightarrow S^1$, choose $x_0\in S^1$ and let $h(x_0) = x_1$. Consider the homomorphism $$h_\ast: \pi_1(S^1,x_0) \rightarrow \pi_1(S^1,x_1).$$

Since both groups are infinite cyclic, we have $$h_\ast(\gamma_{x_0}) = (\gamma_{x_1})^d.$$

I'm trying to show that if two maps $h,k: S^1 \rightarrow S^1$ are homotopic then they have the same degree.

This is a problem from Munkres and I'm unsure where to begin. Honestly a little unclear about the exact definition of degree above too.

Answer 1

In general, if two maps are homotopic, then their induced maps conincide:

For a map $f:(X,x_0)\rightarrow (Y,y_0)$ (that is a continuous map $X\rightarrow Y$ with $f(x_0)=y_0$) we denote its induced map $\pi_1(X,x_0)\rightarrow \pi_1(Y,y_0)$ by $f_*$.

Claim: If $f,g:(X,x_0)\rightarrow (Y,y_0)$ are homotopic relative to $\{x_0\}$, then $f_*=g_*$.

Proof: Choose a homotopy $h:X\times I\rightarrow Y$ of $f,g$ and $[\gamma]\in \pi_1(X,x_0)$. Then $[f\circ \gamma]=[g\circ \gamma]$ by the homotopy $h^\prime:I\times I\rightarrow Y$ given by $h^\prime(s,t)=h(\gamma(s),t)$. $\square$

So if two maps $h,k:S^1\rightarrow S^1$ are homotopic relative to a point, they have the same degree because their induced maps conincide.

Edit:

An analogous statement is true if we look at the induced maps on the first homology: $H_1(X)\rightarrow H_1(Y)$. This leads also to an equivalent definition of the degree of a map $S^1\rightarrow S^1$ and avoids the base point issue.