How does this map fail to be a homotopy?

Question

Consider the map $H: [0,2\pi) \times I \rightarrow \mathbb R^2 \setminus \{0\}, h(x,t)=e^{itx}$ which maps the interval $[0,2\pi)$ to a circular arc whose length grows with $t$, from just one point to a full circle.

Now this map can't be a homotopy, since $S^1$ is not null-homotopic in $\mathbb R^2 \setminus \{0\}$, but don't quite see where exactly $H$ fails to be continuous.

Edit: As a followup after reading Zev's answer, what about the map $G: [0,2\pi] \times I \rightarrow \mathbb R^2 \setminus \{0\}, h(x,t)=e^{itx}$? It can't be a homotopy by the same argument as above, but I still can't see why it wouldn't be continuous. The critical point to consider should obviously be $(1,0) \in \mathbb R^2$, but all open neighbourhoods i've tried to calculate seem to be "ok".

Answer 1

Apologies, my original answer had an error. The answer is still that the map indicated in the question isn't a "homotopy to $S^1$", but I incorrectly stated that replacing $[0,2\pi)$ with $[0,2\pi]$ provided the correct map to consider, when in fact it should have been $S^1$ (of course, any map out of a contractible space is null-homotopic).

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Excellent question! It's good to closely examine where a result conflicts with your intuition, and figure out what's going on. In this case, the map $H$ is a perfectly valid homotopy - it just isn't a "homotopy to $S^1$". That is, the final stage of the homotopy, the map $$H_1:[0,2\pi)\to\mathbb{R}^2\setminus\{0\},$$ is not the map $\phi:S^1\to\mathbb{R}^2\setminus\{0\}$ defined by $\phi(z)=(\mathrm{Re}(z),\mathrm{Im}(z))$ (which is not null-homotopic).

The difference is that, for $0<t_1,t_2<2\pi$ with $t_1>0$ "close to" $0$ and $t_2$ "close to" $2\pi$, continuous maps $S^1\to X$ must send $z_1=e^{it_1}$ and $z_2=e^{it_2}$ to points that are "close together" in $X$, whereas continuous maps $[0,2\pi]\to X$ can send $t_1$ and $t_2$ to any points in the same path component, not necessarily ones that are close together.