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".


1 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).

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.

  • $\begingroup$ Just to check if I understand correctly - in the quotient topology $[0, \varepsilon]$ is not an open set, but in the image of H it is open? And, more generally, the image of a path $\gamma : I \rightarrow X$ with $\gamma(0) = \gamma(1)$ has a different topology than the image of a continous map $\beta : S^1 \rightarrow X$ with the same image in X? $\endgroup$
    – Benno
    Feb 8, 2012 at 17:37
  • $\begingroup$ @Benno: In answer to your second question - no, a continuous map $\gamma$ from $I=[0,1]$ with $\gamma(0)=\gamma(1)$ is exactly equivalent to a continuous map from $S^1$ - it's the half-openness of $[0,2\pi)$ that's the issue. $\endgroup$ Feb 8, 2012 at 17:39
  • $\begingroup$ The quotient topology on $S^1$ from $[0,2\pi)$ is just the topology of $[0,2\pi)$, since no points were identified. In this topology, the points $\gamma(2\pi -\epsilon)$ and $\gamma(0)$ in $S^1$ are not "close", while in the quotient topology from $[0,2\pi]$, they are (since the ends of the interval got "glued together") $\endgroup$ Feb 8, 2012 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.