Showing that identity and g are not homotopic (without Homology)

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.

• Do you find 'they induce different maps on the fundamental group' similarly unsatisfying? – 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.