TITLE: Homotopy equivalences between some sphere-based spaces (quotient of spheres, bouquet of spheres, difference of spheres)
QUESTION [8 upvotes]: I'd like to prove the following equivalences ($k < n$):

$S^n / S^k \sim S^n \vee S^{k+1}$;
$S^n \backslash S^k \sim S^{n-k-1}$.

Low-dimension cases (e.g. $S^2 / S^0$, $S^2 / S^1$, $S^n / S^1$, $S^2 \backslash S^1$) are completely obvious. Is there any way to prove that for arbitrary dimensions?
I have been trying to develop approaches based on mathematical induction (for $k = 0,1,\dots$) and representation spheres as quotient spaces ($S^n \sim D^n / S^{n-1}$). Unfortunately, all these attempts failed for me: I could not achieve any significant simplification bringing me closer to the solution. Any suggestions?

REPLY [3 votes]: Here's a geometric proof of statement $1$. Jim Belk has already given a rigorous proof along with a proof of $2$, but I guess this is worth a shot.

For a given topological space $X$, $\Sigma X$ (called the suspension of $X$) is the space $X \times [1, -1]/\sim$ where $\sim$ identifies the $X \times \{1\}$ copy to a point and the $X\times \{-1\}$ copy to another point.

I presume $S^k$ is embedded in a standard way in $S^n$. If it's the case, $S^n$ can be thought off as $S^k$ suspended $n-k$ times. So first suspend $S^k$ once to get $S^{k+1}$, and suspend that to get $S^{k+2}$, and so on until you finally have $S^n$, with each $S^i$ being the equator of $S^{i+1}$.
$1.$ $S^k$ is the equator inside $S^{k+1} \cong \Sigma(S^k)$, which in turn is the equator of $S^{k+2} \cong \Sigma(\Sigma(S^k))$. $\Sigma(S^k)/S^k$ then is homeomorphic to $S^{k+1} \vee S^{k+1}$. So $\Sigma(\Sigma(S^{k+1}))/S^k$ is the space with the equator replaced appropriately by $S^{k+1} \vee S^{k+1}$. As $S^{k+1} \vee S^{k+1}$ is homotopy equivalent to $S^{k+1}$ with a $(k+1)$-cell pasted at the equator, $\Sigma(\Sigma(S^k))/S^k$ is homotopy equivalent to $S^{k+2}$ with a $(k+2)$-cell attached along a $(k+1)$-sphere on the surface, which is in turn homotopy equivalent to $S^{k+2} \vee S^{k+1}$. Continue this up until $S^n$, so that the resulting quotient $S^n/S^k$ becomes $S^n \vee S^{k+1}$.

Here's a rough sketch of the way you could do this, with the sphere being $S^{k+2}$, with equator $S^{k+1}$ and equator of that in turn is $S^k$. Identifying $S^k$ to a point is the same as attaching a $(k+1)$-disk along the boundary, and slide that $(k+1)$-disk so that the space becomes (modulo homotopy equivalence) $S^{k+2} \vee S^{k+1}$ (the little loop is actually a sketch of $S^{k+1}$). The space $X$, after quotienting, is nothing but $S^n$ with that little loop sticking out.