This is a topology question:
Prove that the ribbon $R := \lbrace (x, y) \in R^2 \, | \, |x| \ge |y|\rbrace$ is homotopy equivalent to the line $\mathbb{R}$.
My thoughts:
Define $$f: R \rightarrow \mathbb{R} , \text{ with } (x,y)\mapsto x$$
Define $$g: \mathbb{R} \rightarrow R , \text{ with } x \mapsto (x,0)$$
Then $f \circ g: \mathbb{R} \rightarrow \mathbb{R}$ as $f \circ g(x) = x = \mathrm{id}_{\mathbb{R}}$, $g \circ f: R \rightarrow R$ as $g \circ f(x) ((x,y)) = g(x) = (x,0)$.
Now define $$F: R \times I \rightarrow R, \text{ with } ((x,y),t)\mapsto (x,ty) \text{ with } (x,y) \in \mathbb{R}, \text{ and } |x| \ge |y|, t \in [0,1] $$
Since $|x| \ge |y| \ge |ty| =t(y)$ and $(x,ty) \in \mathbb{R}$, $F$ is a well defined continuous map.
$F((x,y),0) = (x,0) = g \circ f (x,y)$, $F((x,y),1) = (x,y) = \mathrm{id}_{R} (x,y)$. So $g \circ f$ and $\mathrm{id}_{R}$ are homotopic, and $f \circ g = \mathrm{id}_{\mathbb{R}}$
$F$ is a homotopy of $\mathrm{id}_{S^1}(x,t)$ and $g \circ f (x,t)$. From the definition, I conclude $R$ and $\mathbb{R}$ are homotopy equivalence.
Is this correct? If not, please help me with this?
Thank you in advance for the help!