Let $f : \Bbb R^2 \setminus \{0\} \to S^1$, $f(x) = \frac{x}{\|x\|}$. Show that $f$ is a homotopy equivalence. 
Let $f : \Bbb R^2 \setminus \{0\} \to S^1$, $f(x) = \frac{x}{\|x\|}$. Show that $f$ is a homotopy equivalence.

I was instructed that the inclusion map should be helpful here, but if I let $g : S^1 \to \Bbb R^2 \setminus \{0\} $ be the inclusion then $f \circ g = f(g(x)) = f(x)$ but is this homotopic to $id_{S^1}$? Initially I thought that $f \circ g = id_{S^1}$, but for example $$f(1,1) = (\frac{1}{\sqrt2}, \frac{1}{\sqrt2}) \ne id_{S^1}(1,1)?$$
I’m not sure if $id_{S_1}(1,1)$ is even defined since $(1,1) \notin S^1$?
Also $g \circ f = g(f(x)) = f(x) = id_X$?
What might I be doing wrong here?
 A: In fact your first thought was correct, $f \circ g = id_{S^1}$.
You are right, $id_{S^1}(1,1)$ is undefined. However, that is not an impediment to proving the equation $f \circ g = id_{S^1}$, because the functions $id_{S^1}$, and $g$, and $f \circ g$, all have the same domain, and that domain is $S^1$. Therefore $f \circ g(1,1)$ is also undefined.
I suggest that when you when write out an equation of functions that you need to prove are equal or are homotopic, you very carefully keep track of the domains and ranges of the functions. So, in this case what you are trying to prove is
$$f \circ g = id_{S^1} : S^1 \to S^1
$$
Writing it this way makes it very clear how to prove this... which is almost what you did, but your mathematical grammar needs some improvement. Here's the correct proof:

Given $x \in S^1$, we must prove $f \circ g(x)=id_{S^1}(x)$:
$f \circ g(x) = f(g(x)) = f(x) = x = id_{S^1}(x)$

Comparing this with what you wrote in the last lines, I suggest that you be very careful about using function notation correctly, in particular that you use function arguments correctly. Your proof has two violations of the grammar of function arguments: neither your equation $f \circ g = f(g(x))$ nor your equation $f(x)=id_{S^1}$ makes any sense.
A: Let $x\in S^1$, $\|x\|=1$ implies that $f(x)=x$, it results that $f\circ g(x)=f(x)=x$ and $f\circ g$ is homotopic to $Id_{S^1}$.
Let $x\in \mathbb{R}^2-\{0\}$, $g\circ f(x)=g({x\over{\|x\|}})={x\over{\|x\|}}=f(x)$.
Consider $H_t(x)=tx+(1-t){x\over{\|x\|}}$ defined in $\mathbb{R}^2-\{0\}$, t\in [0,1].
$H_t$ is well defined, $tx+(1-t){x\over{\|x\|}}=0$ implies that $t\|x\|+(1-t)=0$,
$(1-\|x\|)t=1$. This implies that
$t={1\over{1-\|x\|}}$, it $\|x\|<1, t>1$ impossible, $t$ is not defined if $\|x\|=1$ and if $\|x\|>1, t<0$ impossible, this implies that $H_t$ is well defined, $H_0(x)=g\circ f(x)=f(x), H_1(x)=x$, implies that $g\circ f=f $ is homotopic to $Id_{\mathbb{R}^2-\{0\}}$.
