Let $f$ and $g$ be paths in $\mathbb{R}^{2}-\{(0,0)\}$. Show that $f$ is homotopic to $g$.
By the definition of a path, we have $f(x):[0,1]\rightarrow \mathbb{R}^{2}-\{(0,0)\}$ and $g(x):[0,1]\rightarrow \mathbb{R}^{2}-\{(0,0)\}$. We want to find a homotopy $F$ such that $F:[0,1]\times [0,1]\rightarrow \mathbb{R}^{2}-\{(0,0)\}$, and $F(x,0)=f(x)$ ; $F(x,1)=g(x)$. I suspect that the homotopy $F(x,t)=(1-t)f(x)+tg(x)$ won't work because the book gives this hint:
Hint: Show that every path is homotopic to the constant path that sends the entire interval to the path's starting point. Then show that two constant paths are homotopic using the fact that $\mathbb{R}^{2}-\{(0,0)\}$ is path connected.
So my question is, why do we have to do this work when the homotopy I have defined seems to work?
Best Answer
Since $\mathbb{R}^2 - (0, 0)$ is path connected, there exists a continuous path $h: [0, 1] \rightarrow \mathbb{R} - (0, 0)$ such that $h(0) = f(0)$ and $h(1) = g(0)$. Define $F:[0, 1]\times [0, 1] \rightarrow \mathbb{R}^{2}-\{(0,0)\}$ as
$$ F(x, t) = \begin{cases} f((1-3t)x), & t \in [0, \frac{1}{3}] \\ h(3t-1), & t \in [\frac{1}{3}, \frac{2}{3}] \\ g((3t-2)x), & t \in [\frac{2}{3}, 1] \end{cases} $$
where the three cases are continuous and agree at the two overlapping points $t=\frac{1}{3}$ and $t=\frac{2}{3}$. Therefore, $F$ is continuous. Moreover, $F(x, 0) = f(x)$ and $F(x, 1) = g(x)$ for all $x \in \mathbb{R}^2 - (0, 0)$. Thus, $F$ is a desired homotopy between $f$ and $g$.
Note that we cannot define $F$ using $F(x,t)=(1-t)f(x)+tg(x)$ since there may exist $x$ and $t$ such that $F(x, t) = (0, 0)$. Above, we avoid this problem since all three functions $f$, $g$ and $h$ map into $\mathbb{R}^2 - (0, 0)$.