In Tu's book "An Introduction to Manifolds" he defines smooth homotopy as follows.
$M,N$ smooth manifolds, two $C^\infty$ maps $f,g:M\to N$ are smoothly homotopic if there is a $C^\infty$ map $F:M\times \Bbb R\to N$, s.t. $F(x,0)=f(x)$ and $F(x,1)=g(x)$.
My problem is the following example $26.5$: $\Bbb R^2-\{0\}$ and $S^1$ have the same homotopy type.
Let $\iota:S^1\to \Bbb R^2-\{0\}$ be the inclusion map (circle radius $1$).
Let $r:\Bbb R^2-\{0\}\to S^1$ via $r(x)=\dfrac{x}{|x|}$. (ignoring an inverse inclusion map.)
Then $r\circ \iota =\mathrm{id}_{S^1}$. The claim is that $$\iota\circ r:\Bbb R^2-\{0\}\to \Bbb R^2-\{0\} \text{ is homotopic to the identity map.}$$
Define $F:\left(\Bbb R^2-\{0\}\right)\times\Bbb R\to \Bbb R^2-\{0\}$ via $$F(x,t)=(1-t)x+t\frac{x}{|x|}$$
Normally this would work if homotopy was defined with the interval $I=[0,1]$. However with the new definition using $\Bbb R$, picking $t=\dfrac{|x|}{|x|-1}$ maps a circle of radius $|x|\neq 1$ to $0$. So the map isn't defined as it's claimed to be.
Is there another map that would work $\tilde{F}:\left(\Bbb R^2-\{0\}\right)\times\Bbb R\to \Bbb R^2-\{0\}$, $\tilde F(x,t)$ which actually works for all $t\in \Bbb R$ and is smooth.
My best guess so far was to replace $t$ by $|\sin(t)|$ but this fails to be smooth whenever $\sin$ is $0$. I was thinking if I could smooth this out, by replacing $\sin$ with a load of bump functions $\rho_n$ for which $\mathrm{supp}(\rho_n)\subseteq [n-1/2,n+1/2]$, and $\rho_n(n)=1.$ Then possibly define
$$\tilde F(x,t)=\left(1-\sum\limits_{n\in \Bbb Z -\{0\} }\rho_n (t)\right) x+\frac{x}{|x|}\sum\limits_{n\in \Bbb Z -\{0\} }\rho_n (t)$$
Would this work or is there something simpler?
In the version of the book available from Springer Link at least, Tu explicitly addresses this problem and defines a homotopy $$F(x,t) = (1-t)^2 x + t^2 \frac{x}{\|x\|} = \left( (1-t)^2 + \frac{t^2}{\|x\|} \right) x$$ which works for any $t \in \mathbb{R}$.
In any case if you have a smooth homotopy $F : X \times [0,1] \to Y$ between two maps $f,g : X \to Y$, you can always find a smooth homotopy $G : X \times \mathbb{R} \to Y$ such that $G(x,0) = F(x,0)$ and $G(x,1) = F(x,1)$. All you need to do is to find a smooth function $\rho : \mathbb{R} \to [0,1]$ such that $\rho(0) = 0$ and $\rho(1) = 1$. By the smooth Urysohn lemma applied to $(-\infty,0]$ and $[1,+\infty)$, these things exist (thanks to Justin Young for pointing this out). Then you let $G(x,t) = F(x, \rho(t))$.