# Showing that there is a continuous semi-conjugacy from $R_{\alpha}$ to $R_{k \alpha}$.

I'm currently studying the topic of endomorphisms and translations of topological groups within dynamical systems, and came across the following exercise:

Show that for any $$k \in \mathbb{Z}, k \neq 0$$, there is a continuous semi-conjugacy from $$R_{\alpha}$$ to $$R_{k \alpha}$$.

Here, $$R_{\alpha}$$ is the rotation of $$S^1$$ by angle $$2 \pi \alpha$$ for $$\alpha \in \mathbb{R}$$, so:

$$R_\alpha x = x + \alpha \mod 1$$.

I have been struggling with dynamical systems in general, so any help or suggestion regarding this question would be much appreciated.

You need to construct a continuous surjective (not necesserily injective) map $$h$$ from $$S^1$$ to itself such that $$h \circ R_{\alpha} = R_{k\alpha} \circ h$$, that is, transforms an $$\alpha$$ rotation into a $$k$$ times $$\alpha$$ rotation. An intuitive guess for that would be $$h(x) = kx$$ since it scales things up by a factor of $$k$$, and it indeed works:
$$h(R_{\alpha}(x)) = h(x+\alpha) = kx + k\alpha = R_{k\alpha}(kx) = R_{k\alpha}(h(x))$$