Homotopy type of the space $\mathbb{S}^n\times \mathbb{D}^m/\sim$ Let $\mathbb{S}^n$​  denotes the $n$​-sphere and $\mathbb{D}^n$​ denotes the $n$​-disk, i.e., $\{\mathbf{x}\in \mathbb{R}^n:\|\mathbf{x}\|\le 1\}$​. I want to understand the homotopy type of the following space
$$
\mathbb{S}^n\times \mathbb{D}^m/\sim,
$$
where $\mathbb{S}^n\times \partial \left(\mathbb{D}^m\right)\sim \{\star\}$​, where $\star$​ is some point on the boundary of disk. We also assume that $n\le m$.

For example, if we take $m=1$​. Then we have
$$
\mathbb{S}^n\times [-1,1]/\sim
$$
which is nothing but
$$
\mathbb{S}^n\times \{-1\}\sim \{\star_1\}~\text{and}~ \mathbb{S}^n\times \{1\}\sim \{\star_2\} \text{ and finally }, \star_1\sim \star_2.
$$
The above space is homotopic to $\mathbb{S}^{n+1}\vee \mathbb{S}^1$ where $\star_1$ or $\star_2$ is the wedge point. One can see it like this, we are taking the suspension of $\mathbb{S}^n$​ and then identifying the two poles.

In general, one should expect the homotopy type to be $\mathbb{S}^{m+n}\vee\mathbb{S}^{m}$​​.
Any help will be appreciated. Thanks.
 A: Let us call your space $X$.  If you fix a point $p\in S^n$, then the image of $\{p\}\times D^m$ in $X$ is homeomorphic to $S^m$, since you've just collapsed its boundary to a point.  Your space $X$ can then be obtained by attaching an $(m+n)$-cell to this copy of $S^m$ via the map $\partial D^{m+n}\to S^m$ that maps $D^n\times(\partial D^m)\to S^m$ by a constant map and $(\partial D^n)\times D^m\to S^m$ by projection onto the second coordinate followed by the usual quotient map $D^m\to S^m$.  Indeed, this $(m+n)$-cell in $X$ is just the map $D^{m+n}\to X$ you get by composing the quotient map $D^{m+n}\to S^n\times D^m$ with the quotient map $S^n\times D^m\to X$.
So, if this attaching map $\partial D^{m+n}\to S^m$ is nullhomotopic, it follows that its cofiber $X$ is homotopy equivalent to $S^m\vee S^{m+n}$.  Now notice that our copy of $S^m$ in $X$ is actually a retract, since we can map $X$ to $S^m$ by just projecting $S^n\times D^m$ onto the second coordinate.  It follows that we can factor the attaching map $\partial D^{m+n}\to S^m$ as a composition $\partial D^{m+n}\to S^m\to X\to S^m$ where the second two maps are the inclusion and the retraction.  But the composition $\partial D^{m+n}\to S^m\to X$ is nullhomotopic because $X$ is obtained by attaching a cell along $\partial D^{m+n}\to S^m$.  Thus the attaching map $\partial D^{m+n}\to S^m$ is nullhomotopic and $X\simeq S^m\vee S^{m+n}$.
