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I’m learning about homotopy and the fundamental groups, and I’m having a hard time visualizing the transformation. In particular, I can’t see when there is no homotopy between loops.

An example from my class is the following picture (the loops are all confined to $S^1$. I draw them that way just to make clear what paths are being travelled).

enter image description here

I can see how the first loop could be deformed into the second one by doing less and less of the “repeated” portion.

Why can’t we have something similar, for example as shown below, to deform a loop into another loop of the opposite direction? Is the condition of continuity violated somehow?

enter image description here

Apologize for the bad quality of the pics.

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Edit: I’m actually thinking of the fundamental group of the circle $S^1$. My understanding is that the group is comprised of the equivalence classes of loops with a base point $x_0$ (here I take it to be $(1,0)$). In what sense is a loop goes a full circle from $(1,0)$ counterclockwise not the same as a loop goes a full circle from $(1,0)$ clockwise? I don’t see how the procedure in the second picture does not describe a homotopy between the two.

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    $\begingroup$ Notice that 3 is not a loop. You cannot have a homotopy between a loop and a path (which is not a loop). In fact, in 2,3,5, I am assuming you somehow end up back at $x_0$ because otherwise your question wouldnt make sense $\endgroup$ – Pratik Apshinge Nov 18 '20 at 19:02
  • $\begingroup$ The loops are all confined to $S^1$. I draw them that way just to make clear what paths are being travelled. Apologize for the confusion. $\endgroup$ – Josh Ng Nov 18 '20 at 19:13
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    $\begingroup$ Both ends must be at the base point all the time. To go from 1 to 2 you must either temporarily detach the other end or cross through the disk. Can you draw the images half-way between 1 and 2? One or three quarters of the way? Then it may be easier to explain which rule you are violating. $\endgroup$ – Jyrki Lahtonen Nov 18 '20 at 19:43
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    $\begingroup$ Here’s the illustration of how I’d go from 1 to 2 in the second picture: imgur.com/xJuewEB Again I didn’t draw the path confined to $S^1$ just to show clearly how a point would travel along it. For example, for $t = 0.25$, what I have in mind is a point starting at $x_0 = (1,0)$, travels $7/8$ of $S^1$ counterclockwise, then takes the same way back to $x_0$ (in clockwise direction of course). $\endgroup$ – Josh Ng Nov 18 '20 at 20:04
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    $\begingroup$ Thanks for the extra pictures. Now I see that continuity is broken between $t=0$ and $t=0.25$. The loop at $t=0$ ENDS (at $x_0$) after completing one lap. It does not go once around the circle and then trace that same route backwards. $\endgroup$ – Jyrki Lahtonen Nov 19 '20 at 7:19
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The problem is steps $1\to 2$ and $4\to 5$. Perhaps you could imagine this problem as having a rubber band wrapped once around a metal bar (attached to a wall, say, like a ladder rung). Is there any way to get the rubber band off the bar? Or to wrap it around the bar but in the opposite direction?

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  • $\begingroup$ Is it implicitly required that the only times the base point is visited are the beginning and the end of the loop (and not any other time in between)? $\endgroup$ – Josh Ng Nov 18 '20 at 19:30
  • $\begingroup$ No. This comment was too short so here are some more characters $\endgroup$ – vujazzman Nov 18 '20 at 19:31
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It may take a while to identify the origin of the confusion. I can no longer help with that in the space of a comment, so an answer it is.

Here is the first problem that the extra image by the OP revealed. When going from picture 1 to picture 2, they seem to have in mind the homotopy shown in the animation below. Reload the page, if the animation has stopped. I have forgotten how to prepare a gif-file that runs continuously.

enter image description here

And now the problem is apparent. Instead of the picture in 1, one of the endpoints of this homotopy is the path around the circle that first traverses it counterclockwise, and then traces it back clockwise. This is homotopic to the identity element of $\pi_1(S^1,x_0)$, it is not the generator. After all, it is easy to see how to continue this homotopy and eventually shrink the path to the "constant" path at $x_0$.

In yet other words, we have a homotopy between this

enter image description here

and this (the path doing a 180 degree turn at the half point)

enter image description here

but the intended target path

enter image description here

in your figure 1 is nowhere to be seen.

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  • $\begingroup$ Brian M. Scott has more experience in identifying possible sources of confusion in a first course on algebraic topology. If he shows up, we can make siwfter progress. $\endgroup$ – Jyrki Lahtonen Nov 19 '20 at 8:27
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    $\begingroup$ I think your comment about keeping $\alpha = 0.5$ fixed and moving $t$ from $0$ to $1$ helped. In this way I can at least represent the thing in $2$-d. Would it be correct to say that, $H(0.5, 0) = \pi$, while $H(0.5,k) \approx 0$ for $k$ a positive value very close to $0$? So that in a graph with the coordinates $t$ and $H$, as $t$ increases from $0$ we suddenly jump from $(0,\pi)$ to $(k,0)$? $\endgroup$ – Josh Ng Nov 19 '20 at 11:36
  • $\begingroup$ Sorry I’ve had to make things very explicit to see where I got it wrong. Also how did you make that gif? $\endgroup$ – Josh Ng Nov 19 '20 at 11:37
  • $\begingroup$ By the way your animation is exactly what I have in mind. I realize something weird is going on with my deformation, but struggle to see the discontinuity. $\endgroup$ – Josh Ng Nov 19 '20 at 11:39
  • $\begingroup$ @JoshNg In the animation there is no discontinuity. It is just that it does not bridge the gap between your figures number 1 and 2. See the bottom two paths. They are not the same (nor homotopic). $\endgroup$ – Jyrki Lahtonen Nov 19 '20 at 13:11
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Two curves being homotopoc is context-dependent. If you are told that two curve homotopic, you should ask yourself “homotopic where”?

Your second diagram explains why the first and the second circles are homotopic on $\Bbb C$. Yes, they are! But what's important here is that they are not homotopic on $\Bbb C\setminus\{0\}$. Note that when you started to change your circle, you wer unable to avoid passing through $0$.

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  • $\begingroup$ @JoshNg your diagram has left the circle entirely in steps (2), (3), and (4). You are showing a homotopy between two paths in $\mathbb{R}^2$. $\endgroup$ – hunter Nov 18 '20 at 19:21

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