# Strong deformation retraction $\mathbb{I}\!\times\!\mathbb{B}^l \longrightarrow \{1\}\!\times\!\mathbb{B}^l\cup \mathbb{I}\!\times\!\{0\}$?

In Shastri's Elements of Differential Topology, p. 225, there is written:

I don't understand this map $R$. Why is $\theta\!\in\!\mathbb{S}^{l-1}$, shouldn't we have $\theta\!\in\!\mathbb{B}^l$? Furthermore, how do I construct such a map myself? I see that the idea is for each $(s,x)\!\in\!\mathbb{I}\!\times\!\mathbb{B}^l$ to construct a ray starting in $(0,\frac{x}{\|x\|})\!\in\!\{0\}\!\times\!\mathbb{S}^{l-1}$ and going through $(s,x)$, i.e. the ray $$\mathbb{I}\!\times\!\mathbb{B}^l\!\times\![0,\infty)\!\rightarrow\!\mathbb{I}\!\times\!\mathbb{B}^l,\;\; (s,x,t)\mapsto \begin{cases} (0,\frac{x}{\|x\|})+t\big((s,x)-(0,\frac{x}{\|x\|})\big)& x\!\neq\!0\\ (s,x) & x\!=\!0 \end{cases},$$ and then for each $(s,x)$ to choose the right $t_0\!\in\![0,\infty)$ where that ray hits $\{1\}\!\times\!\mathbb{B}^l\cup \mathbb{I}\!\times\!\{0\}$, divide that ray by $t_0$ so that the ray becomes a line parametrized by $\mathbb{I}$. I do not see a way to do this.

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$\mathbb{S}^{l-1}$ is the boundary of the disc $\mathbb{D}^l$. You can consider a point in the disc as a pair $(\theta,s)$, with $0\leq s\leq 1$. – Sigur Aug 16 '12 at 0:23
Ok, so $\theta\!=\!\frac{x}{\|x\|}$. But how did he get the formula for $R$? – Leon Aug 16 '12 at 0:34
Considering $r$ as the time, I guess that for $0\leq r\leq 1/2$, it sends the disc $\{r\}\times \mathbb{D}^l$ to a point on the segment $[0,1]\times \{0\}$ at time $2r$, i.e. $\{2r\}\times \{0\}$, and for the other half $1/2\leq r\leq 1$ it sends $\{r\}\times \mathbb{D}^l$ onto the discs $\{1\}\times \mathbb{D}^l_\rho$, where $\rho$ is related to the radius of the disc (starting with the disc with radius 0, i.e. the point $\{1\}\times \{0\}$ and ending with the disc with radius 1, i.e. $\{1\}\times \mathbb{D}^l$). – Sigur Aug 16 '12 at 0:55
Roughly speaking, during the first half you collapse over the vertical segment and during the last half you project over the top concentric (from the center to the boundary) discs. – Sigur Aug 16 '12 at 0:58
@LeonLampret I have been told that in algebraic topology it is common practice not to write down explicit formulas for deformation retracts/homotopies/etc. For example, if you look in Hatcher many deformation retracts are presented as pictures. For me I have to the point that if you show me the picture, I will accept it as my deformation retract.... – user38268 Aug 16 '12 at 0:59

Claim: $\mathbb{B}^l\!\times\!\mathbb{I}$ strongly deformation retracts onto $\mathbb{B}^l\!\times\!\{1\} \cup \{0\}\!\times\!\mathbb{I}$, and $\mathbb{B}^l\!\times\!\{0\}$ goes to $(0,0)$.
Proof: Each element of $\mathbb{B}^k$ is written in polar coordinates as $(\theta,r)\in\mathbb{S}^{k-1}\!\!\times\!\mathbb{I}$, since $\frac{\mathbb{S}^{k-1}\!\times\mathbb{I}}{\mathbb{S}^{k-1}\!\times\{0\}}\approx\mathbb{B}^k$. The strong deformation retraction $h$ is obtained by not changing $\theta$ (i.e. $h$ is the same on each slice $\{(\theta,r,s)\!\in\!\mathbb{S}^{k-1}\!\!\times\!\mathbb{I}\!\times\!\mathbb{I};\,\theta\!=\!\text{const.}\}\!\approx\!\mathbb{I}\!\times\!\mathbb{I}$), on the lower half of the cylinder, $\mathbb{B}^l\!\times\![0,\frac{1}{2}]$, the outer line $\{(\theta_0,1)\} \!\times\! [0,\frac{1}{2}]$ is pushed to the middle line $\{(\theta_0,0)\} \!\times\! [0,1]$, on the upper half of the cylinder, $\mathbb{B}^l\!\times\![\frac{1}{2},1]$, the outer line $[\frac{1}{2},1] \!\times\!\{(\theta_0,1)\}$ is pushed to the top radial line$\{\theta_0\}\!\times\mathbb{I}\times\!\{1\}$, where the point $(1,\theta_0,1)$ does not move. Let us write $(r,s)$ instead of $(\theta_0,r,s)$. For $0\!\leq\!s\!\leq\!\frac{1}{2}$ the point $(1,s)$ therefore travels to $(0,2s)$, i.e. we have a straight path $t \mapsto (1,s) + t\big((0,2s)\!-\! (1,s)\big) \!=\!\big(1\!-\!t, s(1\!+\!t)\big)$. For $\frac{1}{2}\!\leq\!s\!\leq\!1$ the point $(1,s)$ travels to $(2(s\!-\!\frac{1}{2}),1)$, i.e. we have a path $t \mapsto (1,s) +t\big((2s\!-\!1,1) \!-\!(1,s)\big) \!=\!\big(1\!+\!2t(s\!-\!1),s\!+\!t(1\!-\!s)\big)$. Let us construct the homotopy for the remaining points as well (those from the whole square $\mathbb{I}\!\times\!\mathbb{I}$). The interval $0\!\leq\!s\!\leq\!\frac{1}{2}\!+\!\frac{1-r}{2} \!=\!\frac{2-r}{2}$ will be mapped at the end of the homotopy to $0\!\leq\!s\!\leq\!1$, hence each point $(r,s)$ travels to $(0,\frac{2}{2-r}s)$, i.e. we have a path $t \mapsto (r,s) +t\big((0,\frac{2}{2-r}s)\!-\! (r,s)\big) \!=\!\big(r\!-\!rt,s\!+\!t(\frac{2}{2-r}s\!-\!s)\big) \!=\!\big(r(1\!-\!t),s\!+\!st\frac{r}{2-r}\big)$. Now we must deal with the remaining points $(r,s)$ from the triangle $\frac{2-r}{2}\!\leq\!s\!\leq\!1$. The line through $(0,1)$ and $(1,\frac{1}{2})$ is $y\!-\!1\!=\!\frac{1/2-1}{1-0}(x\!-\!0)$, i.e. $y\!=\!-\frac{1}{2}x\!+\!1$. The line through $(1,1)$ in $(r,s)$ is $y\!-\!1\!=\!\frac{s-1}{r-1}(x\!-\!1)$. These two lines intersect in $(\frac{2s-2}{r+2s-3},\frac{r+s-2}{r+2s-3})$. The ratio of the distances is $\|-(1,1)\!+\!(r,s)\| \,/\, \|-(1,1)\!+\!(\frac{2s-2}{r+2s-3}, \frac{r+s-2}{r+2s-3})\|\!=\!\ldots\!=\!|r\!+\!2s\!-\!3|\!=\!3\!-\!r\!-\!2s\in[0,1]$. Therefore the point $(r,s)$ is mapped to $\big(1\!-\!(3\!-\!r\!-\!2s),1\big)\!=\! (r\!+\!2s\!-\!2,1)$, i.e. we have a path $t\mapsto (r,s)\!+\!t\big( (r\!+\!2s\!-\!2,1) \!-\!(r,s)\big) \!=\!\big(r\!+\!2t(s\!-\!1),s\!+\!t(1\!-\!s)\big)$. The desired strong deformation retraction is therefore
$$\;\;\;\;\;\;\;h\!:\;\;\mathbb{B}^l\!\times\!\mathbb{I}\! \times\!\mathbb{I}\!\longrightarrow\!\mathbb{B}^l\!\times\!\mathbb{I},\;\;\; (\theta,r,s,t)\mapsto \begin{cases} \displaystyle \big(\theta,r(1\!-\!t),s\!+\!st{\textstyle\frac{r}{2-r}})\big) & \hspace{-8pt} ;\;\; \displaystyle r\!+\!2s\!\leq\!2,\\ \displaystyle \big(\theta,r\!+\!2t(s\!-\!1),s\!+\!t(1\!-\!s)\big) &\hspace{-8pt} ;\;\; \displaystyle r\!+\!2s\!\geq\!2. \;\;\;\;\;\;\;\;\;\blacksquare \end{cases}$$