3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Do you find 'they induce different maps on the fundamental group' similarly unsatisfying? $\endgroup$ – user98602 Mar 27 '15 at 16:30
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.