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I am trying to prove the following problem. Show that $\{0\} \times D^n \cup \mathbb{I} \times \mathbb{S}^{n - 1}$ is the deformation retract of $\mathbb{I} \times D^n$. I think I have the visual idea of how to do it, but I am having troubles putting it into a formula. Suppose we consider the point $(2,0) \in R^n \times D^n$, then if we draw rays emanating from that point then we will get our required deformation retract. However, I am having troubles turning it into a formula. Here the precise issue that I am missing. Suppose we have a point $(s,x) \in \mathbb{I} \times D^n$, then how can we determine the end point of the ray which connects (s,x) to (2,0) ? If anyone could help that would be awesome.

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    Can you formulate the region that the endpoint of the ray lies on the bottom disk $\left\{0\right\}\times D^n$? – 2017-02-02
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    How do I know even if a specific point $(s,x)$ will lie in a ray which is connects to bottom disk $\{0\} \times D^n$ ? – 2017-02-02
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    If a point $(s,x)$ is mapped into the bottom disk $(0,y)$, then $(2,0),(s,x),(0,y)$ lie on the same line. So $(s,x)$ must be the convex combination of $(2,0)$ and $(0,y)$, meaning $(s,x)=t(2,0)+(1-t)(0,y)$. Then an elementary school calculus gives you $t=\frac{s}{2}$, thus $x=(1-\frac{s}{2})y$. Can you extend this argument to the complete formula? – 2017-02-02
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    I see I understand this part. Similiarly we can do it for $\mathbb{I} \times S^{n-1}$ that is if $(s,x)$ is mapped into $(m_1,s_1)$, then we must have $t(2,0) + (1 - t)(m_1,s_1) = (s,x)$ where $x \in S^{n - 1}$, so we must have $t = ||x|| - 1$ right ? – 2017-02-02
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    I am little bit confused still how do on how we get a specific formula for the homotopy ? How do we determine if points do classified on $\{0\} \times D^n$ or $I \times S^{n - 1}$ ? – 2017-02-02
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    I guess what you mean is find the region of points which lie on ray which connects to bottom disk $\{0\} \times D^n$ and similiarly for $I \times S^{n - 1}$ ? – 2017-02-02
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    Okay, let me summarize the point. What you did on $I\times S^{n-1}$ is correct, and then the formula for the homotopy is already included in these reasonings. The ray-type homotopy $H$ we want to formulate should map $(s,x)$ to $(0,y)=(0,\frac{2-s}{2}x)$ (in the former case) by constant speed in time 1, thus $H(t,(s,x))=t(0,\frac{2-s}{2}x)+(1-t)(s,x)$. And I'm sure you can do the other part. – 2017-02-02
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    I can do the other part yeah, but how do you determine which parts get we use based on (s,x) ? – 2017-02-02
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    I mean H is really a map from $I \times (I \times D^n)$ into $(I \times D^n)$ how do we determine which map we use for a specific $(s,x)$ ? – 2017-02-02
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    That's what you already did; for example, if your $(s,x)$ are to be mapped into the bottom disk, then $(s,x)$ should satisfies the condition $||x||\leq 1-\frac{s}{2}$, and otherwise you are in the other case. – 2017-02-02
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    oh right right okay cool awesome. Thank you so much. – 2017-02-02
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    It was very valuable discussing with you. Thanks again. – 2017-02-02
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    @cjackal I made a full answer. I will answer my own post in few hours from now, so if this could benefit other people. – 2017-02-02

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