For a homework I have to find the Brouwer degree of all rotations on $S^1$ of this form:
Let $S^1 = \{(x, y) \in \mathbb{R}^2 \mid x^2+y^2= 1\}$ and consider $$\forall k \in \mathbb{Z}: f_k: S^1 \to S^1: (cos 2\pi t,sin 2\pi t) \mapsto (cos 2\pi kt,sin 2\pi kt).$$
I thought I solved this question, but found the correct answer in 2.32 of Algebraic topology. The problem is, I still cannot find the mistake in my answer.
So, this is my answer:
I first use a different representation of $S^1$, namely the one in the complex plane, $f_k: S^1 \to S^1: z \mapsto z^k$. Then I propose a homotopy between $f_k$ and the identity function:
\begin{equation*}
H: S^1 \times I \to S^1: (z, t) \to z^{t(k – 1) + 1}.
\end{equation*}
This map is continuous since the exponential map, '$+$' and '$-$' are all continuous on $\mathbb{C} \setminus \{0\}$. Furthermore, $H(z, 0) = z = Id(z)$ and $H(z, 1) = z^{(k-1) + 1} = z^k$, so all $f_k$ are homotopic to the identity map. And because homotopic maps have the same degree they have Brouwer degree equal to $1$.
So, can you help me find the mistake?
Best Answer
Your homotopy $H$ isn't continuous at the basepoint on the $x$-axis. The exponential map $x \mapsto e^x$ is continuous on $\mathbb{C}$, but the exponential $t \mapsto z^t$ is not continuous in general. To define it in terms of the regular exponential (via $z^t = e^{t\log z}$), you need to choose a branch of the logarithm. The regular branch has a discontinuity on the $x$-axis.