Homotopy equivalence of the two sphere with poles identified Hello I am having trouble with the following exercise:
We are asked to show that the following spaces are homotopic equivalent:

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*The two sphere with north and south pole identified: $\mathbb S^2/\{N ∼ S\}$. (Here N and S represent the north and south pole.


*The two sphere with a interval glueing the north and south pole together: $(\mathbb S^2\cup[0, 1])/\{N ∼ 0, S ∼1\}$.
I need to write down two functions $f,g$ s.t. they are homotopic inverses of each other.
Intuitively it makes sense to me since we can retract the interval to a single point, and the end points of the interval are identified with the north and south pole. And since the interval is contractible it is homotopic equivalent to a single point. But writing explicit functions seems kinda hard.

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*Definition: Two spaces $X, Y$ are homotopic equivalent if there exists $f:X\to Y$, and $g:Y\to X$ s.t. $g\circ f ∼ id_X$ and $f \circ g ∼ id_Y$. The composition of the maps is homotopic to the identity.

We were also given a hint to write the two sphere as two copies of discs with the boundary glued together. I don’t really see how this helps with finding the functions $f,g$.
Can someone give a hint on how to define the functions $f$ and $g$?
 A: Let $$ X= (\mathbb S^2\cup[0, 1])/\{N ∼ 0, S ∼1\},\qquad \qquad Y=\mathbb S^2/\{N ∼ S\}.$$
Then $f\colon X\to Y$ is easy to define.  Following your intuition of retracting the interval to a point, just set $f|_{\mathbb S^2}=1_{\mathbb S^2}$ and $f(t)=N=S$ for all $t\in [0,1]$.
Now we need to define $g\colon Y\to X$.  The advantage of regarding $\mathbb S^2$ as a union of a pair of (unit) disks, is that on each disk we can parameterize using polar co-ordinates, with $S$ or $N$ as the origin.
The idea now is simple: Firstly $g\colon N,S\mapsto 0.5\in [0,1]$.  Then for a point $(r,\theta)$, on either disk of $Y$, if $r\leq0.5$ we map $$g\colon (r,\theta)\mapsto 0.5\pm r\in [0,1].$$
Here the sign $\pm$ is determined by which disk $(r,\theta)$ lies on.
Finally if $r\geq0.5$ we map $$g\colon (r,\theta)\mapsto (2r-1,\theta).$$
Now you have to construct the homotopies $g\circ f ∼ id_X$ and $f \circ g ∼ id_Y$.  Let me know if you need help with that.

Let's start with the composition $f \circ g$.  This collapses points of radius less than $0.5$ to $N=S$, and applies the function $2r-1$ to radii greater than $0.5$.
We need to construct $H_t\colon Y\to Y$ for $t\in [0,1]$ so that $H_0=1_Y$ and $H_1=f \circ g$.
It makes sense then for $H_t$ to collapse points of radius less than $t/2$ to $N=S$.  Then radii from  $t/2$ to $1$ should get mapped to radii from $0$ to $1$, which we can do linearly:
$$H_t(r,\theta)=H_t\left(\left(\frac {r-\frac t2}{1-\frac t2},\theta\right)\right),\qquad r\geq t/2,$$
and $H_t(r,\theta)=N=S$ if $r\leq t/2$.
Now check these two definitions agree at the boundary $r=t/2$, and that $H_0=1_Y$ and $H_1=f \circ g$.  Check also that points of radius $1$ get mapped to points of radius $1$ for all $t$, as this is needed to ensure continuity where the disks are glued.
