What is a full torus divided by its boundary?

Question

I always thought that the quotient space $(S^k\times D^n)\,/\,(S^k\times \partial D^n)$ equals $S^k\times S^n$, at least up to homotopy type. It looks so natural.. but answering this and then reading this, I realized it's simply wrong.

It seems to me that the quotient is some kind of "reduced product" $$ (S^k\times S^n)\,/\,(S^k\times \{*\}) $$

Any hint for a better description, or at least for computing the homotopy groups?

Answer 1

The case $k=n=1$ shows pretty clearly what goes wrong. $S^1 \times S^1$ is the torus, whereas $(S^1 \times D^1) / (S^1 \times \partial D^1)$ is the torus with a single longitude pinched down to a point. That pinching can kill homotopy group elements.

So for example they don't even have the same fundamental group: the torus has fundamental group $\mathbb{Z}^2$, whereas $(S^1 \times D^1) / (S^1 \times \partial D^1)$ has fundamental group $\mathbb{Z}$.

Answer 2

Here are some of the lower homotopy groups. Let $k\geq 1$ and $n\geq 2$. We have the following isomorphisms: $$\pi_q(S^k\times D^n)\cong \pi_q(S^k)\quad\text{and}\quad \pi_q(S^k\times\partial{D^n})\cong\pi_q(S^k)\times\pi_q(S^{n-1}).$$ Therefore $S^k\times D^n$ is $(k-1)$-connected and $S^k\times\partial{D^n}$ is $\min\{k-1,n-2\}$-connected. Furthermore, we can regard $S^k\times\partial{D^n}$ as a subcomplex of $S^k\times D^n$, so that $(S^k\times D^n,S^k\times\partial{D^n},x_0)$ is a CW-pair.

Hence, the projection $p:(S^k\times D^n,S^k\times\partial{D^n})\to((S^k\times D^n)/(S^k\times\partial{D^n}),\ast)$ induces a map $$p_*:\pi_q(S^k\times D^n,S^k\times\partial{D^n},s_0)\to\pi_q\left(\frac{S^k\times D^n}{S^k\times\partial{D^n}},\ast \right)$$ which is an isomorphism for $2\leq q\leq k+1+\min\{k-1,n-2\}$ and an epimorphism for $q=k+1+\min\{k-1,n-2\}+1$. So for $2\leq q\leq k+1+\min\{k-1,n-2\}$, it suffices to compute the relative group.

Now, we consider the piece: $$\pi_q(S^k)\times\pi_q(S^{n-1})\xrightarrow{\ p_{1}\ }\pi_q(S^k)\xrightarrow{}\pi_q(S^k\times{D^n},S^k\times\partial{D^n})\xrightarrow{\ \partial\ }\pi_{q-1}(S^k)\times\pi_{q-1}(S^{n-1})\xrightarrow{p_1}\pi_{q-1}(S^k)$$ of the long exact sequence of the pair $(S^k\times D^n,S^k\times\partial{D^n},s_0)$. The map $p_1$ above is given by the composition: $$\pi_q(S^k)\times\pi_q(S^{n-1})\xrightarrow{\ \cong\ }\pi_q(S^k\times\partial{D^n})\xrightarrow{\ i_* \ }\pi_q(S^k\times D^n)\xrightarrow{\ \cong \ }\pi_q(S^k)\times\pi_q(D^n)\xrightarrow{\ \cong \ }\pi_q(S^k)$$ and following an element through shows that this really is just the projection $$\pi_q(S^k)\times\pi_q(S^{n-1})\to \pi_q(S^k).$$ Exactness then implies that $$\pi_q(S^k\times D^n,S^k\times\partial{D^n},s_0)\cong \ker{p_1}=\pi_{q-1}(S^{n-1}),\quad q\leq k+1+\min\{k-1,n-2\}.$$