Is $S^{n-1}$ a deformation retract of $S^{n}$ \ {$k$ points}?

Question

In this question, they explain that a deformation retract of $S^{^2}\backslash${$k$ points} ($k\geq 2$) is $S^1\vee \ldots\vee S^1$ ($k-1$ copies of $S^1$, joint at a point) by actually describing a deformation retract. I understand all of this.

However, when I first came across this question, my answer was the following:

Assume that none of the missing points lie on the equator and 2 of them are the north and south pole. Consider the deformation retract $r:S^2\backslash${2 points}$\rightarrow S^{1}$ and restrict it to $S^{^2}\backslash${$k$ points}. This restricted map $r$ would still satisfy all properties of a deformation retract, which shows that that $S^1$ is a deformation retract of $S^{^2}\backslash${$k$ points}. But this contradicts that it is $S^1\vee \ldots\vee S^1$, since their first homology group is different.

Where am I wrong? I think it must be that the restricted map $r$ is no longer a deformation retraction, but I don't see why not.

I have asked my question for $S^2$, but a general answer for $S^n$ would be helpfull.

Answer 1

Certainly, the way you are restricting the map is not a homotopy. Look at this way: if X = S^{2}-{ k points} and k>1, you can enlarge the region of a one deleted point and make it flat which is homotopically equivalent to disk minus k-1 points which deformation retract onto k-1 wedge of circles.

Answer 2

It seems that you argue that the restriction of a (strong) deformation retraction $r : X \to A$ to any $Y$ with $A \subset Y \subset X$ is again a (strong) deformation retraction. It is of course a retraction, but in general no longer a homotopy equivalence. A trivial example is $X = [0,1], A = \{ 0 \}, Y= \{0,1\}$.