adjunction space $\mathbf{S}^{n-1}\times\mathbf{D}^{n}\cup\mathbf{D}^{n}\times\mathbf{S}^{n-1}$ is homeomorphic to $\mathbf{S}^{2n-1}$ Consider the adjunction space $\mathbf{S}^{n-1}\times\mathbf{D}^{n}\cup\mathbf{D}^{n}\times\mathbf{S}^{n-1}$ by gluing $\mathbf{S}^{n-1}\times \mathbf{S}^{n-1}$ together, is there a good way to prove that $\mathbf{S}^{n-1}\times\mathbf{D}^{n}\cup\mathbf{D}^{n}\times\mathbf{S}^{n-1}\cong\mathbf{S}^{2n-1}$? When $n=1$, I can check manually that it is true (attach four lines together by glueing their end points, constructing a square that is homeomorphic to $\mathbf{S}^{1}$), but I don't know how to prove it in general.
Thanks!
 A: $S^{n-1} \times D^{n} \cup D^{n} \times S^{n-1}$ is the boundary of $D^n \times D^n$
(by the formula $\partial(M \times N) = \partial M \times N \cup M \times \partial N$)
And since $D^{n} \times D^n$ is homeomorphic to $D^{2n}$ its boundary is $S^{2n-1}$.
Thus we conclude that $S^{n-1} \times D^{n} \cup D^{n} \times S^{n-1} \approx S^{2n-1}$
A: First observe that you are not really gluing something. Both $\mathbf{S}^{n-1}\times\mathbf{D}^{n}$ and $\mathbf{D}^{n}\times\mathbf{S}^{n-1}$ are genuine subspaces of $\mathbf{D}^{n}\times\mathbf{D}^{n}$ and $B := \mathbf{S}^{n-1}\times\mathbf{D}^{n}\cup\mathbf{D}^{n}\times\mathbf{S}^{n-1}$ is their ordinary union. But in fact $\mathbf{S}^{n-1}\times\mathbf{D}^{n}\cap\mathbf{D}^{n}\times\mathbf{S}^{n-1} = \mathbf{S}^{n-1}\times \mathbf{S}^{n-1}$. You can of course regard $B$ as the adjunction space $X \cup_f Y$ with $X = \mathbf{S}^{n-1}\times\mathbf{D}^{n}, Y = \mathbf{D}^{n}\times\mathbf{S}^{n-1}$ along the inclusion map $X \supset \mathbf{S}^{n-1}\times \mathbf{S}^{n-1} \stackrel{f}{\hookrightarrow} Y$. But this is unnecessary here.
Also note that the topologies on $\mathbf{S}^{n-1}$ and $\mathbf{D}^{n}$ are the subspace topologies inherited from $\mathbb R^n$. The topology on $\mathbf{D}^{n}\times\mathbf{D}^{n}$ is the product topology which agrees with the subspace topology inherited from $\mathbb R^{n} \times \mathbb R^{n} = \mathbb R^{2n}$. Thus also $B$ carries the subspace topology inherited from $\mathbb R^{2n}$.
Let $\lVert - \rVert$ denote the Euclidean norm on $\mathbb R^{m}$. On $\mathbb R^{2n} = \mathbb R^{n} \times \mathbb R^{n}$ define
$$\lVert (x,y) \rVert' = \max(\lVert x \rVert, \lVert y \rVert) .$$
This is also a norm on $\mathbb R^{2n}$ and it is equivalent to $\lVert - \rVert$. In fact, all norms on $\mathbb R^{m}$ are equivalent. But note that we do not need this general theorem. We have
$$\max(\lVert x \rVert, \lVert y \rVert) = \max\left(\sqrt{\sum_{i=1}^n x_i^2}, \sqrt{\sum_{i=1}^n y_i^2} \right) \le \sqrt{\sum_{i=1}^n x_i^2 + \sum_{i=1}^n y_i^2} = \lVert (x,y) \rVert,$$
$$\lVert (x,y) \rVert^2=  \sum_{i=1}^n x_i^2 + \sum_{i=1}^n y_i^2 \le 2 \max\left(\sum_{i=1}^n x_i^2, \sum_{i=1}^n y_i^2 \right) = 2 \max(\lVert x \rVert^2,\lVert y \rVert^2 ) \\ = 2 \max(\lVert x \rVert,\lVert y \rVert)^2  = 2 (\lVert (x,y) \rVert')^2 .$$
Let $S  = \{(x,y) \in \mathbb R^{2n} \mid \lVert (x,y) \rVert' = 1\}$ denote the unit sphere with respect to $\lVert - \rVert'$. We have $S = B$. In fact, we have $(x,y) \in S$ iff both $\lVert x \rVert, \lVert y \rVert \le 1$ and at least one of them is $1$, in other words iff $(x,y) \in \mathbf{S}^{n-1}\times\mathbf{D}^{n}$ or $(x,y) \in \mathbf{D}^{n}\times\mathbf{S}^{n-1}$.
Define
$$\phi: S \to \mathbf{S}^{2n-1}, \phi(z) = \frac{z}{\lVert z \rVert } ,$$
$$\psi: \mathbf{S}^{2n-1} \to S, \psi(z) = \frac{z}{\lVert z \rVert' } .$$
These maps are continuous with respect to the standard subspace topologies on $\mathbf{S}^{2n-1}$ and $S$ which are inherited from the "usual" Euclidean topology on $\mathbb R^{2n}$. This topology is by definition induced by the Euclidean norm $\lVert - \rVert$, but is also induced by the norm $\lVert - \rVert'$ because it is equivalent to $\lVert - \rVert$. Thus the continuity of $\phi, \psi$ follows from the fact that $\lVert - \rVert$ and $\lVert - \rVert'$ are continuous real-valued functions on $\mathbb R^{2n}$.
Since $\psi \circ \phi = id$ and $\phi \circ \psi = id$, we see that they are homeomorphisms which are inverse to each other.
