# How to show that $X - \{0\}$ deformation retracts to $Y\$?

Consider the following pushout diagram $$:$$

$$\require{AMScd} \begin{CD} S^{n-1} @>{\iota}>> D^n\\ @V{p}VV @VV{}V\\ Y @>>{}> X\end{CD}$$

Therefore $$X = (D^n \sqcup Y)/ x \sim p(x),$$ $$x \in S^{n-1}.$$ Show that $$X - \{0\}$$ deformation retracts to $$Y.$$

In the lecture note I am following I came across a formula for retract which is as follows $$:$$

Define $$H : (X - \{0\}) × I \longrightarrow X - \{0\}$$ as

$$H(x,t)= {\begin{cases} (1-t)x + t \dfrac {x} {\|x\|}, & x \in D^n - \{0\} \\ x, & x \in Y \end{cases}}$$

But I have few questions here.

First of all why is $$H$$ well-defined? Here $$X - \{0\}$$ is a collection of all equivalence classes. So in order to show that $$H$$ is well defined we need to verify that if $$x \in S^{n-1}$$ then $$x$$ and $$p(x)$$ both map to the same element in $$X - \{0\},$$ which is clear from the definition of $$H,$$ since $$x \sim p(x)$$ in $$X - \{0\},$$ for all $$x \in S^{n-1}.$$

Secondly, why is $$H$$ continuous? I am trying to apply Pasting lemma here. For that we need to show that both $$D^n - \{0\}$$ and $$Y$$ are closed in $$X - \{0\}.$$ But why are they so? I think for that we need $$Y$$ to be compact and $$X$$ to be Hausdorff since continuous image of a compact set is compact and compact subset of a Hausdorff space is closed. Here $$Y$$ is automatically compact since it is the image of $$S^{n-1}$$ under the continuous map $$p$$ and $$S^{n-1}$$ is compact. But I don't have any idea as to why the identification space $$X - \{0\}$$ is Hausdorff.

Finally, why is $$H(x,1) \in Y,$$ if $$x \in Y\$$? It is clear that $$H(x,0) = x,$$ for all $$x \in X - \{0\}$$ and $$H(x,t) = x,$$ for all $$x \in Y$$ and for all $$t \in I.$$ Now for $$x \in Y$$ we have $$H(x,1) = x \in Y.$$ Now if $$x \in D^n - \{0\}$$ then $$H(x,1) = \frac {x} {\|x\|} \in S^{n-1}.$$ But since $$x \sim p(x),$$ for all $$x \in S^{n-1},$$ it follows that $$\frac {x} {\|x\|} \in Y,$$ in the identification space $$X - \{0\}.$$ Is my reasoning correct in this case?

If "yes" then the only part which is left to show is the continuity of $$H.$$ Would anybody please help me in this regard?

• For continuity you could argue that $X-\{0\}$ is the pushout of the same diagram but with $D^n-\{0\}$ and that taking product with $I$ preserves pushouts (since $I$ is compact metric). Then you use the universal property of the pushout to construct the map $H$ (basically what is done in tha notes, i guess), continuity is automatic Apr 27 at 22:02
Your arguments for the first and the third claims look good to me. Following Leonard's suggestion, we think of $$(X-\{0\})\times I$$ as the pushout of $$p\times I:\mathbb{S}^{n-1}\times I\to Y\times I$$ and $$\iota\times I:\mathbb{S}^{n-1}\times I\to (\mathbb{D}^n-\{0\})\times I$$.
The desired map $$H$$ you have written down is then induced by $$H|_{(\mathbb{D}^n-\{0\})\times I}:(\mathbb{D}^n-\{0\})\times I\to X-\{0\}$$ and $$H|_{Y\times I}:Y\times I\to X-\{0\}$$. If both are continuous, it follows from the universal property of pushouts in topological spaces and continuous maps, $$H$$ is also continuous.
The map $$H|_{Y\times I}$$ is just the projection onto the first factor hence continuous. For $$H|_{(\mathbb{D}^n-\{0\})\times I}$$, since addition and multiplication of continuous maps of $$\mathbb{R}^n$$ (with the usual metric topology) are continuous, it remains to show that $$g:\mathbb{D}^n-\{0\}\to \mathbb{D}^n-\{0\}$$ which sends $$\mathbf{x}$$ to $$\mathbf{x}/\|\mathbf{x}\|$$ is continuous. Given $$\mathbf{y}$$, for any $$\mathbf{x}$$ such that $$|\mathbf{x}-\mathbf{y}|<\frac{\varepsilon\|\mathbf{y}\|}{2}$$, we have \begin{align*} |g(\mathbf{x})-g(\mathbf{y})|&=\frac{\left| \|\mathbf{y}\|\mathbf{x}-\|\mathbf{x}\|\mathbf{y}\right|}{\|\mathbf{x}\|\cdot\|\mathbf{y}\|}\\ &=\frac{\left| \|\mathbf{y}\|\mathbf{x}-\|\mathbf{x}\|\mathbf{x}+\|\mathbf{x}\|\mathbf{x}-\|\mathbf{x}\|\mathbf{y}\right|}{\|\mathbf{x}\|\cdot\|\mathbf{y}\|}\\ &\leq \frac{\left| \|\mathbf{y}\|\mathbf{x}-\|\mathbf{x}\|\mathbf{x}\right|+\left|\|\mathbf{x}\|\mathbf{x}-\|\mathbf{x}\|\mathbf{y}\right|}{\|\mathbf{x}\|\cdot\|\mathbf{y}\|}\\ &= \frac{\left| \|\mathbf{y}\|-\|\mathbf{x}\|\right|\cdot\|\mathbf{x}\|+\left|\mathbf{x}-\mathbf{y}\right|\cdot\|\mathbf{x}\|}{\|\mathbf{x}\|\cdot\|\mathbf{y}\|}\\ &\leq \frac{\left| \mathbf{y}-\mathbf{x}\right|\cdot\|\mathbf{x}\|+\left|\mathbf{x}-\mathbf{y}\right|\cdot\|\mathbf{x}\|}{\|\mathbf{x}\|\cdot\|\mathbf{y}\|}<\varepsilon. \end{align*}
• Y is not necessarily open in $X-\{0\}$: consider the case $p=id_{S^{n-1}}$. then $X-{0}=S^{n-1}\times [0,1)$ and $Y=S^{n-1}\times \{0\}$ is not open in it. Apr 27 at 21:56