Prove that $S^n$ (n-dimensional sphere with unit radius) is a strong deformation retract of $\mathbb{R}^{n+1} \setminus\{(0,…,0)\}$
This is my attempt:
Consider $f : S^n \to \mathbb{R}^{n+1} \setminus\{(0,…,0)\}$ as the inclusion map and $g : \mathbb{R}^{n+1} \setminus\{(0,…,0)\} \to S^n$ as the map defined as
$$g(y) = \frac{y}{\operatorname{norm}(y)} \ ,\ y \in \mathbb{R}^{n+1} \setminus\{(0,…,0)\}$$
Clearly both are continuous, and that $g|_{S^n}$ is an identity map, hence $g$ is a retraction.
But when I am faced with proving that $f \circ g$ is homotopic relative $S^n$ to the identity map $1:\mathbb{R}^{n+1} \setminus\{(0,…,0)\} \to \mathbb{R}^{n+1} \setminus\{(0,…,0)\}$, I get stuck.
Are the maps I defined wrong from the start? If they aren't, can someone give me a hint on how to construct the homotopy relative $S^n$? Thank you.
P.S. I'm rather new in Algebraic Topology so there is a lot that I don't know or still do not understand.
Best Answer
So $f:S^n\to\Bbb R^{n+1}∖\{(0,...,0)\}$ is the inclusion and $g:\Bbb R^{n+1}∖\{(0,...,0)\}\to S^n$ is the retraction. We want a homotopy starting with the identity $\text{id}$ end ending at $fg$. Since for any vector $x$ there is a straight line between $x$ and $fg(x)=x/||x||$, one natural way would be the homotopy sliding each point along this line segment, that is the map $$(x,t)\mapsto t\frac x{||x||}+(1-t)x$$ Another map is suggested by the idea that each point $x$ starts at $\dfrac x1$ and ends at $\dfrac x{||x||}$, so in between we should divide $x$ by something between $1$ and $||x||$. This leads to $$(x,t)\mapsto \frac x{t||x||+(1-t)}$$ I leave it to you to check that these maps are well-defined and continuous, then choose your favorite one ;-)