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Two paths are homotopic if one can be continuously deformed to the other, right? So I've been told that the fundamental group of the circle is isomorphic to the integers, since you can't deform e.g., the path that goes around once clockwise to that which goes around twice clockwise, since the disc gets in the way, so it's generated by the the clockwise and counterclockwise paths. But that seems to imply that if you go around once clockwise and once counterclockwise, you get a path that can be continuously deformed to the base point? How?

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As @Ben says, the key is that the middle point, where you switch from finishing the first loop to starting the second loop, is able to move. Here's some bad pictures to feed your intuition a bit.

If you wrap a rubber band once around a post and let go, it will stay around the post:

enter image description here

On the other hand, if you wrap it once, then wrap it the other way, you get this (and when you let go, the band will snap back and have nothing to do with the post):

enter image description here

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The reason one clockwise/counterclockwise loop is not nulhomotopic is that both the beginning and end of the loop must stay fixed. If you were allowed to move either one of the endpoints during the homotopy, then this path would be contractible. If you do one clockwise then one counter clockwise loop, then only the start of the first clockwise loop and end of the second counterclockwise loop must remain fixed, while the middle point is allowed to move. This allows you to contract the whole loop.

Concretely, if $$f(t) = \begin{cases} e^{4\pi i t}&t\leq 1/2\\ e^{-4\pi i t}&1/2\leq t \leq 1 \end{cases}$$

then there is a nulhomotopy $$h(t,s) = \begin{cases}f(ts)&t \leq 1/2\\f(s(1-t)) & 1/2\leq t\end{cases}$$

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    $\begingroup$ Presumably start and beginning should be beginning and end? $\endgroup$ – timtfj Jan 30 '19 at 14:39
  • $\begingroup$ @timtfj Fair enough :) $\endgroup$ – Ben Jan 30 '19 at 15:45
  • $\begingroup$ The $h$ you provide is not nullhomotopic. Indeed, check the endpoints are fixed for all values of $s$. they are correct for $0$ and $1$ but nowhere in between. What you describe is a line that makes it $\frac{1}{s}$ of the way around the path that is one loop one way, one loop back. The problems are: you wish the condition ($t$ less/greater than $0.5$) to remain true of $t$, not the function argument, and, the function will not be smooth at switch in the current form. $\endgroup$ – drjpizzle Jan 31 '19 at 11:15
  • $\begingroup$ @drjpizzle Yes I can see it's not right - I think $f(st)$ should be correct for $t \leq 1/2$, but not when $t \geq 1/2$. $\endgroup$ – Ben Jan 31 '19 at 11:19
  • $\begingroup$ Well it's smooth and you have a nulhomotopy exists so you can extend it to one... I have put the formula as a comment to my answer (in response to yours.) $\endgroup$ – drjpizzle Jan 31 '19 at 11:36
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I think this is best answered directly/constructively:

Going a fraction $k$ about the circle one way and them coming back again, as $k$ varies smoothly between $0$ and $1$ should do the trick.

At $k = 0$ this is the point. At $k = 1$ this is one around and once back. I think its clear it's continuious.

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  • $\begingroup$ Why it is right in the general case (the winding number arguments given) is a useful result and nice answers. It's the right way to think about it, and understand why this is a nice result. But to the particular question answered, I think this is the most direct answer. $\endgroup$ – drjpizzle Jan 30 '19 at 20:30
  • $\begingroup$ By the way this is what the nulhomtopy $h(t,s)$ in a previous answer does, if you plug in $k$ (in this answer) in the $s$ variable. $\endgroup$ – Ben Jan 31 '19 at 10:44
  • $\begingroup$ @Ben This is not the case. In your example, take $f(st)$ for $s = 0.5$. This becomes: $e^{2 \pi i t}$ or one full loop. Not half a loop forward and half a loop back. I know what you are getting at but I think you made a mistake. I will comment on your answer to see if i can make this clear. $\endgroup$ – drjpizzle Jan 31 '19 at 11:06
  • $\begingroup$ Thanks you're right. I'll try to fix the formula for $h$, if you have any suggestions please let me know. $\endgroup$ – Ben Jan 31 '19 at 11:14
  • $\begingroup$ @Ben I think in this case, a wordy answer might be best. But: $h(t,s) = e^{2s ((2 \pi) t i)}$ when $t < 0.5$, $e^{2s ((2 \pi) (1 - t) i)}$ otherwise, I think is correct. edit: grammar and silly mistakes. $\endgroup$ – drjpizzle Jan 31 '19 at 11:25

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