$S^n$ is a retract of $B^{n+1}\setminus{P}$ I would like to prove that $S^n$ is a retract of $B^{n+1}\setminus{P}$ where $P$ is a non-empty subset of $B^{n+1}$ such that $P\cap S^{n}=\emptyset.$
So let $s\in P$ and define $f$ the continuous function that maps $(M,t)\to (1-t)s+tM$ from $S^n\times I\to B^{n+1}\setminus{s}.$
Now, if $\overline{f}:S^n\times (0,1]\to B^{n+1}$ is a homeomorphism then if I denote $\pi: S^n\times (0,1]\to S^n$ the function defined by $\pi((P,s))=P$ then $\pi\circ f^{-1}$ is a retract of $B^{n+1}\setminus{s}$ on $S^n.$
So I just need to prove that $\overline{f}$ is a homeomorphism but not sure how can I proceed.
 A: Notice that your map $\bar{f}$ can also be interpreted as a map 
$$\bar{f}: B^{n+1} \setminus 0 \rightarrow B^{n+1} \setminus s$$ For any $x \in B^{n+1} \setminus 0$, write $x$ uniquely as $tM$ for $M \in S^n = \partial B^{n+1}$ and $t \in (0,1]$ and define $\bar{f}(tM) = (1-t)s + tM$. 
There's an analogous unique decomposition (see below for why it is unique) of any point $x \in B^{n+1} \setminus s$ as the sum of $s$ and a line segment radiating from $s$, i.e. $s + t(M - s)$ for some $M \in S^n$ and $t \in (0,1]$.  But note that $s + t(M-s) = (1-t)s + tM = \bar{f}(tM)$, hence $\bar{f}$ is surjective.  The proof that this decomposition is unique is simultaneously the proof that $\bar{f}$ is injective:  if $(1-t)s + tM = (1-v)s + vN$, then we have $(v-t)s = tM - vN$ and


*

*if $v = t$ then $M = N$

*otherwise $|s| = \frac{|tM - vN|}{|t - v|} \geq  \frac{|tM - vM|}{|t - v|} = 1$ (inequality follows from triangle inequality) which is a contradiction.


So $\bar{f}$ is a bijection.  
We want to show it's a homeomorphism, so let's look now at continuity.  Since we're working with Euclidean balls, we'll work with epsilon-delta definition.  Assume $s$ is not the center of the ball, because in that case $\bar{f}$ was the identity map to begin with and trivially a homeomorphism.  
Consider that for any $x = tM, y = vN$, $$|\bar{f}(x) - \bar{f}(y)| = |(v-t)s + tM - vN| \leq |v-t||s| + |x-y| \leq (1 + |s|)|x-y|$$ 
(The first inequality is triangle, and the second one is the fact that the difference in magnitudes of the points is at most the distance between them).
This means that to keep $|\bar{f}(x) - \bar{f}(y)| < \epsilon$, all we have to do is keep $|x -y| < \frac{\epsilon}{1 + |s|}$ 
That shows continuity of $\bar{f}$.  To show continuity of $\bar{f}^{-1}$, use the unique decomposition discussed earlier to write it explicitly as $\bar{f}^{-1}\big(s + t(M-s)\big) = tM$ and attack it again with epsilon-delta. In particular you have for any $x = s + t(M-s), y= s + v(N-s)$ that $$|\bar{f}^{-1}(x) - \bar{f}^{-1}(y)| = |tM - vN| = |t(M-s) + s - v(N-s) - s + ts - vs| \leq |x - y| + |t-v||s|$$
Now you also have that $$|x -y| = |t(M-s) - v(N-s)| \geq |(t-v)(M-s)| \geq (1 - |s|)|t-v|$$ (distance is minimized when $M, N$ are parallel.  Substituting into the above we see that $$|\bar{f}^{-1}(x) - \bar{f}^{-1}(y)| \leq \frac{|x-y|}{1 - |s|}$$ which shows continuity of $\bar{f}^{-1}$.
As an aside, a useful result that follows from the above established homeomorphism (see the comments below for more explanation) is that $B^n \setminus P_1$ is homeomorphic to $B^n \setminus P_2$ for any sets of interior points $P_1, P_2$ with the same finite cardinality.  From there we can also see that $R^n$ with $k$ points removed is homeomorphic to $R^n$ with any other $k$ points removed, and the same goes for $S^n$.  
