Question on contractability of a hemisphere I was going through some text in algebraic topology.
A text from a book reads:
The punctured sphere $\Bbb{S}^n$ - {point} is homeomorphic to $\Bbb{R}^n$.( I am convinced with this.)
Next it says that:
Therefore, a hemisphere is contractible.
I didn't get this. I know that $R^n$ and any convex subspace of it is contactible. But from here, how can I conclude that a hemisphere is contractible?
 A: Hint:  A hemisphere is a subset of the punctured sphere (since it misses a point of the sphere).  Moreover, it is a retract of the punctured sphere.
A: Southern hemisphere of $\mathbb S^n$ is homeomorphic to closed unit ball in $\mathbb R^n$ (under restriction of stereographic projection, projection taken  onto the plane  $x_{n+1}=0$, not $x_{n+1}=-1$), which being convex is contractible and lower hemisphere is homeomorphic to upper hemisphere under $(x_1,...,x_{n+1}) \mapsto (x_1,...,-(x_{n+1}))$.
Also, the image of hemisphere $\{x \in \mathbb S^n | x_{n+1} \geq 0\}$ under the homeomorphism $(x_1,...,x_{n+1}) \mapsto (x_1,...,0)$, with inverse $(x_1,...,0) \mapsto (x_1,...,\sqrt{1-(x_1^2+...+x_n^2)})$, is also homeomorphic to the closed unit ball in $\mathbb R^n$
A: You do not say whether hemispheres are open or closed. Let us assume they are closed (but everything works also for open hemispheres). Your text states

The punctured sphere $S^n \setminus \{point\}$ is homeomorphic to $\mathbb R^n$. Therefore, a hemisphere is contractible.

The purpose of this argument seems to be to identify a hemisphere $H$ with a certain subset of $\mathbb R^n$ which should be known as obviously contractible. But this argument is deficient. Without knowing anything about a concrete homeomorphism $h : S^n \setminus \{point\} \to \mathbb R^n$, we cannnot say anything about $h(H) \approx H$.
As Akash Gaur observes in his answer, we can take $h=$ stereographic projection and get $h(H)$ = closed unit ball in $\mathbb R^n$ which is obviously contractible. But using this is much more than $S^n \setminus \{point\} \approx \mathbb R^n$.
Anyway, if the goal is to prove that hemispheres are contractible, we can do it much more directly.
A general hemisphere can be described as follows. Let $v \in S^n$ be any point. Then the orthogonal complement $E(v) = \{ x\in \mathbb R^{n+1} \mid \langle v, x \rangle = 0\}$ is a hyperplane and the sets $S^n_+(v) = \{ x\in S^n \mid \langle v, x \rangle \ge 0\}$, $S^n_-v) = \{ x\in S^n \mid \langle v, x \rangle \le 0\}$ are the two closed hemispheres lying in the two halfspaces determined by $E(v)$. Note that $S^n_-(v) = S^n_+(-v)$. Thus it suffices to show that each $S^n_+(v)$ is contractible. Define
$$\phi : S^n_+(v) \times I \to S^n_+(v), \phi(x,t)  = \dfrac{tv+(1-t)x}{\lVert tv+(1-t)x \rVert} .$$
This is well-defined:

*

*Let $\psi(x,t)  = tv+(1-t)x$. We have  $\langle v,\psi(x,t)\rangle = t\langle v,v\rangle + (1-t)\langle v,x\rangle \ge t\langle v,v\rangle = t$. Recall $\langle v,x\rangle\ge 0$ for $x \in S^n_+(v)$.


*We have $\psi(x,t) \ne 0$: This is obvious for $t = 0$. For $t > 0$ it follows from 1.


*The quotient $\phi(x,t) = \dfrac{\psi(x,t)}{\lVert \psi(x,t) \rVert} \in S^n$ is well-defined by 2. Using 1. we get $\langle v, \phi(x,t) \rangle = \dfrac{\langle v, \psi(x,t) \rangle}{\lVert \psi(x,t) \rVert} \ge 0$. Thus $\phi(x,t) \in S^n_+(v)$.
We have $\phi(x,0) = x$ and $\phi(x,1) = v$. Thus the homotopy $\phi$ is a contraction of $S^n_+(v)$ to the point $v$.
Remark:
In the special case of the "upper hemisphere" $H = \{(x_1,\ldots,x_{n+1}) \mid x_{n+1} \ge 0\}$ we can define $h : H \to D^n = \{x \in \mathbb R^n  \mid \lVert x \rVert \le 1\}, h(x_1,\ldots,x_{n+1}) = (x_1,\ldots,x_n)$. This is a homeomorphism with inverse $h^{-1}(y) = (y,\sqrt{1-\lVert y \rVert^2})$. But $D^n$ is obviously contractible. This argument did not use stereographic projection  or some other homeomorphism $S^n \setminus \{point\} \approx \mathbb R^n$.
