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I am trying to show that the mapping cone of $f:S^{1}\rightarrow S^{1}$ defined by $$f(z)=\begin{cases} z^4&\text{for }z\text{ in the upper semicircle}\\ \bar{z}^{2}&\text{for }z\text{ in the lower semicircle} \end{cases}$$ is contractible by showing that $f$ is homotopic to the identity. What is the homotopy?

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(If you've done things with $\pi_1(S^1)$ yet...) The map you have is the concatentation of a degree 2 map and a degree -1 map, and so has degree 1. –  Thomas Belulovich Apr 29 '12 at 17:46
    
As $z$ goes half-way around the circle, $f(z)$ goes around the whole circle twice. Then as $z$ traverses the second half, $f(z)$ goes backwards around the circle once. Then $z$ and $f(z)$ have each gone around once. –  Michael Hardy Apr 29 '12 at 18:41
    
@Justin: (Sorry, I don't seem to have enough reputation to comment on your answer or vote up your answer, which I would gladly do!) Thank you for answering. I understand that $f(e^{2\pi it})=e^{2\pi i(4t)} $for $0 \leq t\leq 1/2$. But why is that $f(e^{2\pi it})=e^{2\pi i(3-2t)}$ for $1/2\le t\leq 1$? If $z=e^{2\pi it}$, then $\bar{z}^{2}=e^{2\pi i(2-2t)}$, no? Therefore, it should be $f(e^{2\pi it})=e^{2\pi i(2-2t)}$ for $1/2\le t\leq 1$. Also, how did you construct the homotopy $H$? Some kind of reparametrization lemma? –  John Apr 30 '12 at 15:24
    
@John $e^{2i \pi}=1$, so $e^{2i\pi(2-2t)} = e^{2i\pi(2-2t)} * e^{2i\pi} = e^{2i\pi(3-2t)}$. –  mercio Apr 30 '12 at 15:52
    
@Justin: Got it, thanks! I guess I need to take a complex analysis course. How did you construct $H$? –  John Apr 30 '12 at 15:59

2 Answers 2

We can identify your map $f$ as $f(e^{2\pi i t}) = e^{2\pi i (4t)}$ for $0\le t\le 1/2$ and $f(e^{2\pi i t}) = e^{2\pi i (3 - 2t)}$ for $1/2 \le t \le 1$. Then, we can use a linear homotopy $H: S^1 \times I \to S^1$ given by $H(e^{2\pi i t}, s) = e^{2\pi i ((4t)(1-s) + ts)}$ for $0\le t \le 1/2$ and $H(e^{2\pi i t}, s) = e^{2\pi i ((3-2t)(1-s) + ts)}$ for $1/2\le t \le 1$. It is easy to check that this map is well defined, and that it satisfies $H(1, s) = 1$. This provides a basepoint preserving homotopy $f\simeq 1$, and thus the mapping cone of $f$ is contractible.

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Justin's answer is more conceptual than this, which is in my opinion the way to go. But I have not covered that material yet. After all this question arises from problem 5 in section 15 of Chapter I in Topology and Geometry, by G.E. Bredon, and there is no mention in that section of covering spaces or lifting of maps. So here is my hands on dirty answer:

Using the notation $z=e^{2\pi it}$ the map becomes $$f(e^{2\pi it})=\begin{cases} e^{2\pi i (4t)} & \text{for $\;0\le t\le 1/2$}\\ e^{2\pi i(2-2t)} & \text{for $\;1/2\le t\le 1$}. \end{cases}$$

I tried constructing what seemed to be a very natural homotopy from $f$ to the identity map on $S^{1}$. That is, the one obtained via the diagram consisting of a square and two line segments. The left joining $(1/4,0)$ and $(1,1)$, and the right joining $(1/2,0)$ and $(1,1)$.

diagram describing a homotopy

However, I was trying to define the homotopy so that it was constant on those line segments. That was the problem. After realizing that, the matter of finding a formula was simple. $$F(e^{2\pi i t},s)=\begin{cases} \exp\left(2\pi i \dfrac{4t}{3s+1}\right) & \text{for $\;0\le t\le \frac{3s+1}{4}$}\\\\ \exp(2\pi i(4t-3s)) & \text{for $\;\frac{3s+1}{4}\le t\le \frac{s+1}{2}$}\\\\ \exp(2\pi i (2-2t)) & \text{for $\;\frac{s+1}{2}\le t\le 1$}. \end{cases}$$

This homotopy is constant on the left line segment, but not on the right. The idea was to notice that in the second and third pieces of $F$ the time we spend travelling along $f$ is proportional to $(1-s)$. For example, this means that, in order to define the second piece of $F$ we are better off by defining a bijective linear map $$\left[\dfrac{3s+1}{4},\frac{s+1}{2}\right]\rightarrow [0,1-s]$$ sending $\tfrac{3s+1}{4}$ to $0$ and the other endpoint to $1-s$.

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The motivation for this exercise is the following result: If two maps are homotopic, their mapping cylinders (and hence their mapping cones) are homotopy equivalent. Therefore, one can show that the dunce cap is contractible because it is easily seen to be homeomorphic to the mapping cone of $f$. –  John May 3 '12 at 4:22

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