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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?

Thanks in advance.

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  • $\begingroup$ 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 $\endgroup$
    – Leonard
    Apr 27 at 22:02
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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.

attachingtoX-0

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*}

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  • $\begingroup$ 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. $\endgroup$
    – Leonard
    Apr 27 at 21:56

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