Homotopy equivalences between some sphere-based spaces (quotient of spheres, bouquet of spheres, difference of spheres) 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?
 A: 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.  
A: The following two theorems are proven in chapter 0 of Hatcher's Algebraic Topology.
Theorem 1. If $X$ is a CW complex and $A$ is a contractible subcomplex, then $X/A$ is homotopy equivalent to $X$.
Theorem 2. Let $X$ be a CW complex, and let $A$ be a subcomplex of $X$.  If $Y$ is a topological space and $f,g\colon A\to Y$ are homotopic maps, then $X\cup_f Y$ is homotopy equivalent to $X\cup_g Y$
Here is a nice corollary
Corollary. Let $X$ be a CW complex, and let $A$ be a subcomplex. Then


*

*$X/A$ is homotopy equivalent to $X\cup_A CA$, where $CA$ is the cone on $A$.

*If the inclusion map $A\to X$ is nullhomotopic, then $X/A$ is homotopy equivalent to $X\lor SA$, where $SA$ is the suspension of $A$.
Proof: For (1), observe that $CA$ is a contractible subcomplex of $X\cup_A CA$. By Theorem 1, it follows that $X\cup_A CA$ is homotopy equivalent to $(X\cup_A CA)/CA$, which is homeomorphic to $X/A$.
For (2), let $f\colon A\to X$ be the inclusion, and let $g\colon A\to X$ be a constant map.  Then $f$ and $g$ are homotopic, so $CA \cup_f X$ is homotopy equivalent to $CA \cup_g X$ by Theorem 2.  The former is $X\cup_A CA$, and the latter is $X\lor SA$.$\quad\square$
It follows that $S^n/S^k$ is homotopy equivalent to $S^n \lor S(S^k) = S^n\lor S^{k+1}$ whenever $k<n$, where $S^k$ denotes any $k$-sphere that can be expressed as a subcomplex of $S^n$ with respect to some CW structure.  This proves statement (1).
For statement (2), recall that $S^n$ is homeomorphic to the join $S^k * S^{n-k-1}$, where $S^k$ is the canonical copy of $S^k$ in $S^n$.  Then $S^n - S^k$ deformation retracts onto $S^{n-k-1}$ in an obvious way.  Note that this only proves the statement for the canonical copy of $S^k$ in $S^n$ -- it's not true for arbitrary copies of of $S^k$.  For example, the complement of a knotted circle in $S^3$ is not homotopy equivalent to $S^1$.
