Question: Are the identity mapping on $S^1$ and the reflection about the $x$-axe homotopic?

This is a question which I already know the answer. The objective is to find better answers and suggestions to improve mine.

Here some definitions for improving the context of the problem.

Notation: let us denote $[0,1]$ by $I$.

Definition: Let $X$ and $Y$ be topological spaces and let $A$ be a subspace of $X$. Let $f,g:X\to Y$ be continuous functions. We say that $f$ is homotopic to $g$ relative to $A$ (denote $f\simeq_A g$) if there is a continuous function $H:X\times I\to Y$ such that the functions of the form $H_t:X\to Y$, $H_t(x)=H(x,t)$ for $t\in I$, satisfy the following:

  1. $H_t(x)=g(x)=f(x)$ for all $t\in I$ and all $x\in A$.
  2. $H_0=f$ and $H_1=g$.

The map $H$ is called an homotopy from $f$ to $g$ relative to $A$. When $f\simeq_\emptyset g$ we say that $f$ and $g$ are homotopic and denote $f\simeq g$. Notice that $\simeq_A$ is an equivalence relation between continuous functions from $X$ to $Y$.

Definition: Two loops $\alpha,\beta:I\to X$ at $x\in X$ ($\alpha$ and $\beta$ continuous s.t. $\alpha(0)=\alpha(1)=\beta(0)=\beta(1)=x$) have the same homotopy class (denoted by $[\alpha]=[\beta]$) iff $\alpha\simeq_{\{0,1\}} \beta$.

Problem: Consider the function $g:S^1\to S^1$, $g(x,y)=(x,-y)$ and $f=id_{S^1}$ the identity map in $S^1$. Prove without using Homology Theory that $g$ and $id_{S^1}$ are not homotopic.

  • $\begingroup$ Do you find 'they induce different maps on the fundamental group' similarly unsatisfying? $\endgroup$ – user98602 Mar 27 '15 at 16:30

We will prove that $f$ and $g$ are not homotopic by contradiction. Let $\alpha:I\to S^1$ be the loop at $p=(1,0)\in S^1$ defined by $\alpha(s)=(\cos(2\pi s),\sin(2\pi s))$. Then the loop $\alpha^{-1}:I\to S^1$, $\alpha^{-1}(s)=\alpha(1-s)$ is such that $[\alpha]^{-1}=[\alpha^{-1}]$. Notice that $[\alpha]$ generates $\pi_1(S^1,p)$.

Suppose that $f\simeq g$. We will show that this implies that $\alpha\simeq_{\{0,1\}}\alpha^{-1}$ or equivalently that $[\alpha]=[\alpha]^{-1}$, which is going to lead us to a contradiction.

Remark: notice that if $f\simeq g$, then $\alpha=f\circ\alpha\simeq g\circ\alpha=\alpha^{-1}$. But the relation $\alpha\simeq\alpha^{-1}$ is not enough to prove $\alpha\simeq_{\{0,1\}}\alpha^{-1}$. Actually, it can be proved that $\alpha\simeq_{\{0\}}\alpha^n$ for all $n\in\mathbb{Z}$, whenever $f$ and $g$ are homotopic or not. For example the function $T:I\times I\to S^1$, $T(s,t)=(\cos(2\pi st),\sin(2\pi st))$ is a homotopy relative to $\{0\}$ from $C_p$ to $\alpha$, where $C_p:I\to S^1$, $C_p(s)=p$, for all $s\in I$.

Let $H:S^1\times I\to S^1$ be a homotopy from $f$ to $g$. Then the continuous mapping $$G:I\times I\to S^1, G(s,t)=H(\alpha(s),t)$$ is such that $G_0=\alpha$, $G_1=\alpha^{-1}$ and $G_t(0)=G_t(1)$ for all $t\in I$. Notice that $G$ is not necessarily a homotopy from $\alpha$ to $\alpha^{-1}$ relative to $\{0,1\}$, because we could have $t_1\neq t_2$ such that $G(i,t_1)\neq G(i,t_2)$ for $i\in \{0,1\}$.

Let $\beta:I\to S^1$ be the loop at $p$ defined by $\beta(t)=G(0,t)$. Since $G$ is continuous, so is $\beta$. Now consider $S^1$ as a subspace of $\mathbb{C}$. Define the function $F:I\times I\to S^1$ as $$F(s,t)=G(s,t)\cdot e^{-i\cdot \arg(\beta(t))}$$ where $\cdot$ is the product of complex numbers and $\arg:S^1\to [0,2\pi]$ is the argument function.

Geometric interpretation of $F$: For $t\in I$ fixed, the function $z\mapsto z\cdot e^{-i\cdot \arg(\beta(t))}$ is a rotation function of angle $-\arg(\beta(t))$.

This function rotates the image of the loop $s \mapsto G_t(s)$ in such a way that the base point of the resultant loop is $p$. In fact, the base point of the loop $s \mapsto G_t(s)$ is $G_t(0)$, and the base point of the resultant loop $s \mapsto G_t(s)\cdot e^{-i\cdot \arg(\beta(t))}$ is $$F(0,t)=G_t(0)\cdot e^{-i\cdot \arg(\beta(t))}=\beta(t)\cdot e^{-i\cdot \arg(\beta(t))}=e^{i\cdot \arg(\beta(t))}\cdot e^{-i\cdot \arg(\beta(t))}=p, \mbox{ for all }t\in I.$$

Is not difficult to check that the function $t\mapsto e^{-i\cdot \arg(\beta(t))}$ is continuous and so is $F$.

Properties of $F$:

$F(s,0)=G(s,0)\cdot e^{-i\cdot \arg(\beta(0))}=G(s,0)=\alpha(s)$

$F(s,1)=G(s,1)\cdot e^{-i\cdot \arg(\beta(1))}=G(s,1)=\alpha^{-1}(s)$

$F(0,t)=p$, as we proved above.

$F(1,t)=G_t(1)\cdot e^{-i\cdot \arg(\beta(t))}=\beta(t)\cdot e^{-i\cdot \arg(\beta(t))}=e^{i\cdot \arg(\beta(t))}\cdot e^{-i\cdot \arg(\beta(t))}=p$

We have proved that $F$ is a homotopy from $\alpha$ to $\alpha^{-1}$ relative to $\{0,1\}$. Therefore $[\alpha]=[\alpha]^{-1}$, then $[\alpha]^2=[1]$. Hence $[\alpha]$ generates a finite group which contradicts the fact that $\pi_1(S^1,p)\cong\mathbb{Z}$.

Therefore $f$ and $g$ are not homotopic.


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