Nulhomotopic map from $S^1 \rightarrow \mathbb{C} - \{0\}$ Hullo, I am aware that the inclusion map $i : S^1 \rightarrow \mathbb{C} - \\{0\\}$ is not nulhomotopic since there is a retraction from $\mathbb{C} - \\{0\\}$ to $S^1$ making the induced homomorphism $i_*$ injective, hence nontrivial. What about the following map between the above two spaces: $e^{2\pi iz} \mapsto e^{4\pi iz}$. Is this map nulhomotopic? Could the same argument work? Couldn't we think of this map as moving at twice the speed, i.e., going around twice in the same time? Thanks. W.

No. The squaring map you write, i.e. $z\mapsto z^2$ if you write $S^1\subseteq\Bbb C$, has degree $2$, so cannot be null-homotopic since it is the composition of the inclusion with the squaring map, i.e. your map decomposes as

$$S^1\stackrel{i}{\longrightarrow} \Bbb C\setminus\\{0\\}\stackrel{z\mapsto z^2}\longrightarrow \Bbb C\setminus\\{0\\}$$

This induces the $\times 2$ map on $\Bbb Z$, hence it's not null-homotopic.