It is a well known fact that the closed disk $\mathbb{D}^n$ minus for example its center, deformation retracts onto the border $\mathbb{S}^{n-1}$, for example this argument can be generalized to the $n$ dimensional case Retraction of disk minus a point.
I was wondering whether in a $CW$ setup, there is a similar fact. Specifically I'm asking whether is possible to write explicitly a deformation retract of the $n+1$ skeleton $X^{n+1}$ minus the center of $(n+1)$-cells on the $n-$ skeleton $X^n$.
I think the idea should use the fact that we know an explicit retraction of the disk minus a point on its border, but I'm not sure how to well define the maps even just to prove the continuity, since we have identifications given by characteristic maps, and maybe family of maps to deal with, which are not so intuitive to me.
I need this "fact" in order to prove that $(X,X^n)$ is $n-$connected if $X$ is a $CW$ complex, any hint,help solution or reference would be appreciated.
Edit : I thought to use the deformation retraction given by $H(x,t) = (1-t)x+t\frac{x}{\lvert\lvert x \rvert \rvert}$ for every disk $\mathbb{D}_\alpha^n$ with center $\vec{0}_\alpha$ and costruct the following diagram $$\begin{array}{cc}
\,\,\,\,\,\,\,X^n – \bigsqcup_\alpha\phi_{\alpha}(\vec{0_\alpha}) \\
\\
\ \pi \uparrow & \,\,\,\,\,\,\,\searrow \,{\simeq} \\
\\
\,\,\,\,\,\,\left( \bigsqcup_\alpha \mathbb{D}_\alpha^n – \phi_{\alpha}(\vec{0_\alpha}) \right) \sqcup X^{n-1} & \xrightarrow{r} & X^{n-1}
\end{array}$$
Where $r$ is defined through the deformation retracts as $\begin{cases} \varphi_{\alpha}\circ r_{\alpha}(x) & x \not\in X^{n-1} \\ x & x \in X^{n-1} \end{cases}$
Issues : Is $r$ continuos or even well defined? Taking the equivalence relation on $\left( \bigsqcup_\alpha \mathbb{D}_\alpha^n – \phi_{\alpha}(\vec{0_\alpha}) \right) \sqcup X^{n-1}$ really yields to $X^n – \bigsqcup_\alpha\phi_{\alpha}(\vec{0_\alpha})?$ Last but not least, I don't if this map induces the homotopy equivalence $\simeq$ on the right of the diagram I'm hoping for.
Best Answer
Let $X^{n+1}$ be the $n+1$-skeleton and $\Phi_{\alpha}\colon D_{\alpha}^{n+1}\rightarrow X^{n+1},\,\alpha\in A$ the $n+1$-cells. Then $X^{n+1}$ with the centers of the $n+1$-cells removed is the space $X^{n+1}\setminus\{\Phi_{\alpha}(0)\colon\alpha\in A\}$. This is the disjoint union of its subspaces $X^n=X^{n+1}\setminus\{\Phi_{\alpha}(\mathrm{int}(D_{\alpha}^{n+1}))\colon\alpha\in A\}$ (here, I'm identifying $X^n$ with its copy in $X^{n+1}$ as is usual) and $\Phi_{\alpha}(\mathrm{int}(D_{\alpha}^{n+1})\setminus\{0\}),\,\alpha\in A$. Let $H\colon(D^{n+1}\setminus\{0\})\times I\rightarrow D^{n+1}\setminus\{0\}$ be a strong deformation retraction of $D^{n+1}\setminus\{0\}$ onto $S^n$. Now, we want to construct a strong deformation retraction $\tilde{H}\colon(X^{n+1}\setminus\{\Phi_{\alpha}(0)\colon\alpha\in A\})\times I\rightarrow X^{n+1}\setminus\{\Phi_{\alpha}(0)\colon\alpha\in A\}$ of $X^{n+1}\setminus\{\Phi_{\alpha}(0)\colon\alpha\in A\}$ onto $X^n$. The idea is to push forward the homotopy $H$ through the $\Phi_{\alpha},\,\alpha\in A$ while leaving $X^n$ fixed, though some care needs to be taken at the boundaries.
Let us define $\tilde{H}(x,t)=x$ for $(x,t)\in X^n\times I$ and $\tilde{H}(x,t)=\Phi_{\alpha}(H(\Phi_{\alpha}^{-1}(x),t))$ for $(x,t)\in\Phi_{\alpha}(\mathrm{int}(D_{\alpha}^{n+1})\setminus\{0\})\times I,\,\alpha\in A$ (note that $\Phi_{\alpha}\vert_{\mathrm{int}(D_{\alpha}^{n+1})}\colon\mathrm{int}(D_{\alpha}^{n+1})\rightarrow X^{n+1}$ is a homeomorphism onto its image, whose inverse is what I mean by $\Phi_{\alpha}^{-1}$). By construction, $\tilde{H}(-,0)$ is the identity, $\tilde{H}(x,t)=x$ for any $(x,t)\in X^n\times I$ and $\tilde{H}(-,1)$ is a retraction of $X^{n+1}\setminus\{\Phi_{\alpha}(0)\colon\alpha\in A\}$ onto $X^n$ (since $\tilde{H}(\Phi_{\alpha}(\mathrm{int}(D_{\alpha}^{n+1})\setminus\{0\})\times\{1\})\subseteq\Phi_{\alpha}(S^n)\subseteq X^n$ for all $\alpha\in A$). It remains to check that $\tilde{H}$ is continuous.
By definition of the CW topology, there is a quotient map $X^n\sqcup\bigsqcup_{\alpha\in A}D_{\alpha}^{n+1}\rightarrow X^{n+1}$. Since $I$ is compact, $\left(X^n\sqcup\bigsqcup_{\alpha\in A}D_{\alpha}^{n+1}\right)\times I\rightarrow X^{n+1}\times I$ is a quotient map. Furthermore, $\left(X^n\sqcup\bigsqcup_{\alpha\in A}D_{\alpha}^{n+1}\right)\times I\cong(X^n\times I)\sqcup\bigsqcup_{\alpha\in A}(D_{\alpha}^{n+1}\times I)$ via the obvious map. Altogether, we obtain a quotient map $(X^n\times I)\sqcup\bigsqcup_{\alpha\in A}(D_{\alpha}^{n+1}\times I)\rightarrow X^{n+1}\times I$, which restricts to a quotient map $r\colon(X^n\times I)\sqcup\bigsqcup_{\alpha\in A}((D_{\alpha}^{n+1}\setminus\{0\})\times I)\rightarrow(X^{n+1}\setminus\{\Phi_{\alpha}(0)\colon\alpha\in A\})\times I$. Chasing the definitions, this is given by $r(x,t)=(x,t)$ for $(x,t)\in X^n\times I$ and $r(x,t)=(\Phi_{\alpha}(x),t)$ for $(x,t)\in(D_{\alpha}^{n+1}\setminus\{0\})\times I$. Since $r$ is a quotient map, it suffices to check continuity of $\tilde{H}\circ r$. By definition of the coproduct topology, it suffices to check the continuity of $\tilde{H}\circ r\vert_{X^n\times I}$ and $\tilde{H}\circ r\vert_{D_{\alpha}^{n+1}\setminus\{0\}\times I}$ for all $\alpha\in A$. For $(x,t)\in X^n\times I$, we have $\tilde{H}(r(x,t))=\tilde{H}(x,t)=x$, whence $\tilde{H}\circ r\vert_{X^n\times I}$ is simply the composition $X^n\times I\rightarrow X^n\rightarrow X^{n+1}$ of projection followed by inclusion, i.e. continuous. For $(x,t)\in(D_{\alpha}^{n+1}\setminus\{0\})\times I$, we have $$\tilde{H}(r(x,t))=\tilde{H}(\Phi_{\alpha}(x),t)=\left.\begin{cases}\Phi_{\alpha}(x),&x\in S_{\alpha}^n,\\\Phi_{\alpha}(H(x,t)),&x\in\mathrm{int}(D^{n+1})\setminus\{0\}\end{cases}\right\}=\Phi_{\alpha}(H(x,t)).$$ In the last equality, we have used that $H$ is a strong deformation retraction. Thus, $\tilde{H}\circ r\vert_{(D_{\alpha}^{n+1}\setminus\{0\})\times I}=\Phi_{\alpha}\circ H$ is continuous. In conclusion, $\tilde{H}$ is continuous and thus a strong deformation retraction.