----- > [!proposition] Proposition. ([[complement of spheres is homotopy equivalent to another one]]) > For $m<n$, consider $\mathbb{S}^{m}$ as a [[subspace topology|subspace]] of $\mathbb{S}^{n}$ given by $\{ (x_{1}, x_{2},\dots, x_{m+1}, 0, \dots,0) : \sum x_{i}^{2} = 1 \}.$ > Then the complement $\mathbb{S}^{n}-\mathbb{S}^{m}$ is [[homotopy equivalent]] to $\mathbb{S}^{n-m-1}$. ![[CleanShot 2024-06-03 at [email protected]]] ^proposition > [!proof]- Proof. ([[complement of spheres is homotopy equivalent to another one]]) > It will suffice to show that that $\mathbb{S}^{n} - \mathbb{S}^{m}$ (strongly) [[deformation retract|deformation retracts]] onto a space $\mathbb{S}^{n-m-1}_{*}$ [[homeomorphism]] $\mathbb{S}^{n-m-1}$. > The prototype to keep in mind is that cutting the equator ($\mathbb{S}^{1}$) off $\mathbb{S}^{2}$ yields two hemispheres which can be 'shrunk' to their respective poles (i.e., $\mathbb{S}^{0}$), and that if we puncture antipodal points ($\mathbb{S}^{0}$) on $\mathbb{S}^{2}$ we may stretch the resulting holes larger and large until just the equator ($\mathbb{S}^{1}$) is left. > Now, consider the set $\mathbb{S}^{n-m-1}_{*}:= \left\{ (\overbrace{0,\dots,0}^{m+1 \text{ 0s}}, x_{m+2}, \dots, x_{n}) : \sum x_{i} ^{2} = 1 \right\}.$ It is [[homeomorphism]] to $\mathbb{S}^{n-m-1} \subset \mathbb{S}^{m}$ are described in the problem definition (as witnessed e.g. by [[permutation matrix|coordinate permutations]], which are in general [[linear isomorphism|linear isomorphisms]]). It is also disjoint from $\mathbb{S}^{m}$, for the two sets (neither of which contain $\boldsymbol 0$) live in [[linear subspace|linear subspaces]] which are [[orthogonal complement|orthogonal complements]] of one another. > Also consider the 'normalized projection map' $\mathcal{P}:\mathbb{S}^{m} - \mathbb{S}^{n} \to \mathbb{S}_{*}^{n-m-1}$ given by $\mathcal{P}(\boldsymbol x)=\mathcal{P}\big( x_{1},\dots,x_{n} \big):= \frac{(\overbrace{0,\dots,0}^{m+1 \text{ 0s} }, x_{m+2}, \dots, x_{n})}{\|(0,\dots,0, x_{m+2}, \dots, x_{n})\|}$ which is [[well-defined]] because (supposing not) if $\boldsymbol x \in \mathbb{S}^{m}$ satisfies $x_{m+2}=\dots=x_{n}=0$ then necessarily $\boldsymbol x \in \mathbb{S}^{n}$. $\mathcal{P}$ is clearly [[continuous]] and in fact a [[retract]] of $\mathbb{S}^{m}-\mathbb{S}^{n}$ onto $\mathbb{S}^{n-m-1}_{*}$. The claim is that $\iota \circ \mathcal{P}$ is [[homotopy|homotopic]] to the [[identity map|identity map]] $\id_{\mathbb{S}^{m}- \mathbb{S}^{n}}$ [[homotopy relative to a subset|relative]] to $\mathbb{S}^{n-m-1}_{*}$. Define $\begin{align} H : (\mathbb{S}^{m} - \mathbb{S}^{n}) \times I \to & \mathbb{S}^{n-m-1}_{*} \\ H(\boldsymbol x, t) := & \frac{(1-t)\boldsymbol x + t\mathcal{P}(\boldsymbol x)}{\|(1-t) \boldsymbol x + t\mathcal{P}(\boldsymbol x)\|}. \end{align}$ We see that $H$ is [[well-defined]] () and [[continuous]] as a ratio of sums/products of [[continuous]] functions, and [[well-defined]] (), and moreover that $\begin{align} H(\boldsymbol x, 0)= & \boldsymbol x / \|\boldsymbol x\| = \boldsymbol x \\ H(\boldsymbol x, 1) = & \mathcal{P}(\boldsymbol x) / \|\mathcal{P}(\boldsymbol x)\| = \mathcal{P}(\boldsymbol x), \end{align}$ hence $H$ is a [[homotopy]] between $\iota \circ \mathcal{P}$ and $\id_{X}$. Actually, since for all $\boldsymbol x \in \mathbb{S}^{n-m-1}_{*}$ and all $t \in I$ it is clear that (since $\mathcal{P}(\boldsymbol x)=\boldsymbol x$) $H(\boldsymbol x , t)=\frac{\boldsymbol x}{\|\boldsymbol x\|}=\boldsymbol x$, $H$ is a [[homotopy]] relative to $\mathbb{S}^{n-m-1}_{*}$. Therefore, $H$ is a (strong) [[deformation retract|deformation retraction]] of $\mathbb{S}^{n}-\mathbb{S}^{m}$ onto $\mathbb{S}^{n-m-1}_{*} \cong \mathbb{S}^{n-m-1}$. Since [[homotopy equivalent|homotopy equivalence]] is an [[equivalence relation]], this completes the proof. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` #reformatrevisebatch02