Intersection of the sphere with the first octant is homeomorphic to the ball I'm trying to show that $B$, which is the intersection of the n-sphere $S^n$ with the nonnegative octant of $\mathbb{R}^{n+1}$ is homeomorphic to the ball $B^n$. 
I see how to do this when $n=2$. First, the projection map of $B$ onto the first octant is a homeomorphism. Then stretching the quarter of a circle to a half circle is also a homeomorphism. And Stretching each vertical line of the half circle to a full circle is also a homeomorphism. I can easily write down an explicit formula for each of these steps.
However, in the general $n$ case, how can I construct an explicit homeomorphism? I would greatly appreciate any help.
 A: $$ B= \{ (x_1,\cdots,x_{n+1} )\in {\bf R}^{n+1} | \sum_i x_i^2 =1,\
x_i\geq 0 \} $$
So we have $$ f: B\rightarrow B',\ f(x_1,\cdots,x_{n+1} ) =
(x_1,\cdots, x_n,0) $$
Fix $p=(\epsilon,\cdots,\epsilon,0)\in B'$ Define $$ C:=
\{(x_1,\cdots, x_n,0) \in B'| \sum_i x_i^2=1 \} $$
Consider $$ c(t)= p + (v-p)t ,\ v\in C $$
Then $$ f(t):= |c(t)|^2 = \sum_i  ( \epsilon +(x_i - \epsilon )t )
^2
$$
So $$ f' = \sum_i 2(  \epsilon +(x_i - \epsilon )t ) (x_i - \epsilon
) = \sum_i 2\epsilon  (x_i - \epsilon ) + 2 (x_i - \epsilon )^2t  $$
$$ f'(t_0)=0,\ t_0=-\frac{\sum_i \epsilon  (x_i-\epsilon )}{ \sum_i
 (x_i-\epsilon)^2 } = \frac{-v\cdot p + |p|^2}{|v-p|^2}  $$
Since $\angle (v,p) < \frac{\pi}{2} $ so $-v\cdot (1,\cdots, 1,0) <
0 $. Since $\epsilon$ is small then $t_0<0$ Hence $f(t)$ is
increasing on $[0,1]$. So $c(t)$ is in $B'$. Note that from this we
have an deformation retract from $B'$ to $p$. Hence we have an
homeomorphism from unit ball to $B'$
A: Hint: You can project $B$ straight down into the $x$-$y$ plane to obtain a closed quarter-disk. Do you see that this is a homeomorphism, and that the closed quarter-disk is homeomorphic to the closed disk (your "balls" must be closed balls in this context)?
