What topological space is this quotient of $D^2\times S^2$?

Question

Let $X=D^2\times S^2$ be the cartesian product of the closed two disk with the two dimensional sphere. Define the equivalence realtion $$ (x,y,\sigma_1)\sim(x,y,\sigma_2) $$ for every $(x,y)\in\partial D^2$ and $\sigma_1,\sigma_2\in S^2$. What is the name (if there is one) of the space $Y=X/\sim$? I've managed to prove that $Y$ has no boundary by finding an open neighbourhood at each point $[x,y,\sigma]$, $(x,y)\in\partial D^2$ by setting $$ U=\{[x',y',\sigma']:\sigma'\in S^2,(x',y')\text{ in an open neighbourhood of }(x,y)\text{ in }D^2\}\subset Y. $$ Then the preimage by the quotient map $f:X\to Y$ is open (recall $D^2$ has boundary). Do you have any insight? In the low dimensional case $D^1\times S^1/\sim$ where we collaps $S^1$ at $\partial D^1=\{-1\}\cup\{1\}$, it is clear that $Y$ is homeomorphic to $S^2$

Answer 1

Fix $n,m\in\Bbb N$ and define the map: $$q:D^n\times S^m\to S^{n+m}$$By: $$(x,y)\mapsto\left(x_1,\cdots,x_n,y_0\sqrt{1-\|x\|^2},y_1\sqrt{1-\|x\|^2},\cdots,y_m\sqrt{1-\|x\|^2}\right)$$Viewing all sides as subsets of Euclidean space. Observe that this map is well defined.

It is clearly a continuous surjection. For compactness reasons it must then be a quotient map.

You can also see that the relation imposed by $q$ is exactly the relation you want for your space $Y$. Therefore: $$D^n\times S^m/((x,\sigma)\sim(x,\sigma')\text{ if }x\in\partial D^n)\cong S^{n+m}$$Is witnessed by $q$. In particular, the commented conjecture about $D^2\times S^2/_\sim\cong S^4$ is correct.

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Following the suggestion of Ben in the comments, consider the following situation. We have a general topological space $X$, a strongly locally compact Hausdorff space $Y$. If $CX\cong X\times I/(X\times\{0\})$, write $q':X\times I\twoheadrightarrow CX$ for the quotient map. We claim that: $$Z:=CX\times Y/(\forall y\in Y,\,(x,y)\sim(x,y')\text{ if }x\in q'(X\times\{1\}))$$Is homeomorphic to the join $X\ast Y$, which can be defined as the quotient of $X\times Y\times I$ by the relations $(x,y,0)\sim(x',y,0)$ for all $x,x'\in X,\,y\in Y$ and $(x,y,1)\sim(x,y',1)$ for all $x\in X,\,y,y'\in Y$. Write $q'':CX\times Y\twoheadrightarrow Z$ for the quotient map.

Recall that $q'(X\times\{1\})\subset CX$ is usually identified with $X$, so we could abuse notation and write this as $(x,y)\sim(x,y')$ if $x\in X$.

There is a simple argument for this: using the fact that the functor $(-)\times Y$ preserves quotients, we have a composite quotient map: $$q:X\times Y\times I\cong(X\times I)\times Y\overset{q'\times1}{\twoheadrightarrow}CX\times Y\overset{q''}{\twoheadrightarrow}Z$$

What is the relation induced by $q$? Suppose $q(x,y,t)=q(x',y',t')$.

From: $q''(q'(x,t),y)=q''(q'(x',t'),y')$, we can conclude that $q'(x,t)=q'(x',t')$ and $y=y'$ or that $q'(x,t)=q'(x',t')\in q'(X\times\{1\})$. In both cases, $q'(x,t)=q'(x',t')$ is forced and in the latter case, $t=t'=1$ is forced.

In the former case, $q'(x,t)=q'(x',t')$ could mean $x=x'$, $t=t'$ (in which case, $(x,y,t)=(x',y',t'))$ or it could mean that $t=t'=0$, so $(x,0,y)$ has been identified with $(x',0,y)$.

In the latter case, $q'(x,1)=q'(x',1)$ means $x=x'$, so that $(x,y,1)$ has been identified with $(x,y',1)$.

Overall, we find that the relation imposed by $q$ is precisely the join quotient relation, so that $Z\cong X\ast Y$ as desired.

Oh, and this abstractly generalises the quotient you asked about since $D^n\cong CS^{n-1}$ (with $\partial D^n$ being the image of $S^{n-1}\times\{1\}$) and $S^i\ast S^j\cong S^{i+j+1}$.