homotopy and induced homomorphism.

Question

In my homotopy lectures, we defined degree function as it follows:

$deg:\pi_1(\mathbb S^1,1)\to\mathbb Z$ determined by the following rule:

Let $f:\mathbb S^1 \to \mathbb S^1$ be a cts. map. Consider the induced homomorphism $f_*:\pi_1(\mathbb S^1,1)\to\pi_1(\mathbb S^1,1)$.

Since $\pi_1(\mathbb S^1,1)$ is an infinite cyclic group, there exists a unique integer $d$ such that $f_*[\alpha]=[f(\alpha)]=d[\alpha]$ for any $[\alpha]\in \pi_1(\mathbb S^1,1)$.

This integer $d$ is called the degree of $f$.

Can anyone help me how to show that $f$ homotopic $g$ iff $deg(f)=deg(g)$, where $f,g:\mathbb S^1 \to \mathbb S^1$ are continuous maps.

Answer 1

We could use the fact that $\mathbb R$ is a cover of $S^1$. One should think of $\mathbb R$ as a "winding staircase" above $S^1$, with a projection map, $$ p: \mathbb R \to S^1, \ \ \ p(x) = (\cos (2\pi x), \sin (2\pi x)).$$

Our first step would be to show that, given any map $\gamma : [0,1] \to S^1$ and any $x_0$ such that $\gamma(0) = (\cos(2\pi x_0), \sin(2\pi x_0))$, there exists a unique "upstairs lift", $\tilde \gamma : [0,1] \to \mathbb R$, with $p \circ \tilde \gamma = \gamma$ and $\tilde \gamma(0) = x_0$.

If we think of $S^1 $ as $[0,1]$ with the endpoints identified, then we can also use this lifting property to lift maps $\gamma : S^1 \to S^1$ to maps $\tilde \gamma : [0,1] \to \mathbb R$. For any such lift, it is clear that $$\tilde \gamma(1) - \tilde \gamma(0) \in \mathbb Z,$$ since $\tilde \gamma(0)$ and $\tilde \gamma(1)$ should project down to the same point $\gamma(0) = \gamma(1)$ on the circle downstairs.

A very useful observation is that the degree of $\gamma$ is precisely this integer: $$ deg(\gamma) = \tilde \gamma(1) - \tilde \gamma(0).$$ Said another way, the degree of the map $\gamma : S^1 \to S^1$ is the number of flights of stairs you climb on the staircase. It should be routine to check that this definition agrees with the definition of the degree that you gave.

So why is this useful? It is useful because if $\gamma_1$ and $\gamma_2$ have the same degree, then we can find lifts $\tilde \gamma_1$ and $\tilde \gamma_2$ in such a way that $$ \tilde \gamma_1 (0) = \tilde \gamma_2 (0), \ \ \tilde \gamma_1(1) = \tilde \gamma_2(1).$$ And since $\mathbb R$ is convex, we can define a homotopy between $\tilde \gamma_1$ and $\tilde \gamma_2$ "upstairs"! In other words we can find an $\tilde F : [0,1] \times [0,1] \to \mathbb R$ such that $$\tilde F(t,0) = \tilde \gamma_1(t), \ \ \tilde F(t,1) = \tilde \gamma_2(t),$$ and crucially, this homotopy can be arranged so that at no point during the process of deforming $\tilde \gamma_1$ into $\tilde \gamma_2$ do we move the endpoints: $$ \tilde F(0,s) = \tilde \gamma_1(0) = \tilde \gamma_2(0), \ \ \tilde F(1,s) = \tilde \gamma_1(1) = \tilde \gamma_2(1) .$$ This guarantees that at no point during the deformation do we break the image of the circle downstairs. Notice how vital it is that $\gamma_1$ and $\gamma_2$ have the same degree, in order to achieve this "non-breaking" property!

Finally, to get the desired homotopy between $\gamma_1$ and $\gamma_2$, we "project" $\tilde F$ downstairs, i.e. we take $p \circ \tilde F$ as our homotopy between $\gamma_1$ and $\gamma_2$. And we're done.

For the "only if" part, we are given a homotopy $F$ between $\gamma_1$ and $\gamma_2$. We could try to show that $\gamma_1$ and $\gamma_2$ can be lifted to $\tilde \gamma_1$ and $\tilde \gamma_2$ as before, and also, the homotopy $F$ can be lifted to a homotopy $\tilde F$ between $\tilde \gamma_1$ and $\tilde \gamma_2$, with $p \circ \tilde F = F$. But then, since $$ \tilde F(0,0) = \gamma_1(0), \ \ \tilde F(0,1) = \gamma_2(0), \ \ \tilde F(1,0) = \gamma_1(1) , \tilde F(1,1) = \gamma_2(1), $$ and since $$ F(0,s) - F(0,0) \in \mathbb Z, \ \ F(1,s) - F(1,s) \in \mathbb Z,$$ (which is to say that we must never break the circles downstairs during our deformation), a continuity argument implies that $$ \tilde \gamma_1 (0) = \tilde \gamma_2 (0), \ \ \tilde \gamma_1(1) = \tilde \gamma_2(1) $$ which proves that $deg (\gamma_1) = deg (\gamma_2)$.

Answer 2

I would like to point out that albeit $\pi_1(\mathbb{S}^1,1)$ is a cyclic group, this does not imply that these elements are integers, as you suggest. However, there is an isomorphism between $\pi_1(\mathbb{S}^1,1)$ and $\mathbb{Z}$.

I only give hints as this is a good exercise to do by yourself. Let us proceed by double implications.

For the direct implication, try to prove the following:

Lemma. Let $f\colon\mathbb{S}^1\rightarrow\mathbb{S}^1$ and $g\colon\mathbb{S}^1\rightarrow\mathbb{S}^1$ be two continuous maps such that for all $x\in\mathbb{S}^1$, one has: $$|f(x)-g(x)|<2.$$ Then, $f$ and $g$ have the same degree.

Sketch of a proof. Consider two continuous lifts $\theta_f\colon\mathbb{R}\rightarrow\mathbb{R}$ and $\theta_g\colon\mathbb{R}\rightarrow\mathbb{R}$ of $f$ and $g$ through the universal covering map $p\colon\mathbb{R}\rightarrow\mathbb{S}^1,t\mapsto e^{2i\pi t}$ such that $|\theta_f(0)-\theta_g(0)|<1/2$ (why can you find such lifts?) and use the intermediate value theorem to prove that this inequality holds also for $t=1$. Conclude recalling the degree of a map is precisely the difference between the value at $1$ and $0$ of one of its lifts. $\Box$

The content of this lemma is that if two maps on the circle are never antipodal, then they make the same number of oriented turns.

Then, applied the lemma to a homotopy $H\colon\mathbb{S}^1\times[0,1]\rightarrow\mathbb{S}^1$ using the uniform continuity of $t\mapsto H(\cdot,t)$. Notice that $C^0(\mathbb{S}^1,\mathbb{S}^1)$ is as always in algebraic topology equipped with the compact-open topology, namely in this case the topology of the uniform convergence.

If you want, you can also use the homotopy lifting property for covering space.

For the reverse implication, you can explicitly write a homotopy with fixed extremities between two maps of the same degree, once again using lifts through the universal covering map of $\mathbb{S}^1$.