$S^n/S^k$ is homotopy equivalent to $S^n\vee S^{k+1}$.

Question

How can I prove that $ S^{n}/S^{k} $ is homotopy equivalent to $S^{n} \vee S^{k+1} $? Here $S^{n} \vee S^{k+1} $ is a space $ \left[ S^{n} \cup S^{k+1} \right]/\approx $ where $ \approx$ is an equivalence relation gluing one specific point from $ S^{n} $ with one specific point from $ S^{k+1} $.

Answer 1

Let us consider $S^n$ as a subspace of $\Bbb R^{n+1}$, and $S^k$ as its subspace which is the unit sphere in $\Bbb R^{k+1}\times\{0\}^{n-k}$. Let $S^n\cup D^{k+1}$ be the union of this $S^n$ with the $(k+1)$-dimensional unit ball in $\Bbb R^{k+1}\times\{0\}^{n-k}$. Since $(S^n,S^{k})$ is a cofibered pair, so is $(S^n\cup D^{k+1}, D^{k+1})$, and since $D^{k+1}$ is contractible, the quotient map $S^n\cup D^{k+1} \to \left( S^n \cup D^{k+1} \right) / D^{k+1} \approx S^n/S^k$ is a homotopy equivalence (see Hatcher's Algebraic Topology, proposition 0.17).

Now $S^n\cup D^{k+1}$ is the space obtained by gluing $D^{k+1}$ to $S^n$ via the map $f:\partial D^{k+1} \to S^n$ which embedds $S^k$ into $S^n$. This map is homotopic to a constant map $g: \partial D^{k+1} \to S^n$. Since furthermore $\left( D^{k+1}, \partial D^{k+1} \right)$ is cofibered, this implies $S^n\cup D^{k+1}$ is homotopy equivalent to $S^n\cup_g D^{k+1}$ (see Hatcher, proposition 0.18), which is the same as $S^n\vee S^{k+1}$.

Also compare this with example 0.14 in Hatcher's book.