A contractible mapping cone?

Question

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?

Answer 1

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 $(1,1)$.

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 $1-s$.

Answer 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.