Are Gamma_4 and Gamma_5 Truly Homotopic?

[ehrenfest]

[SOLVED] homotopic curves

Homework Statement

Apparently if $\gamma_4 = \gamma_2 +\gamma_3 -\gamma_1-\gamma_3$, then $\gamma_4$ is homotopic to $\gamma_5$ in any region containing $\gamma_1$,$\gamma_2$, and the region between them minus z.

I am not convinced that this is true.

I can picture how you could transform $\gamma_4$ into $\gamma_5$ one another by first moving the two $\gamma_3$s apart, then kind of going around the circle and finally contracting the curve, but I am not convinced that this is a continuous function from the unit interval cross the unit interval. Specifically, I am not convinced that you can just move the two $\gamma_3$s apart in a continuous way. How can you rigorously show that there is a homotopy?

Homework Equations

The Attempt at a Solution

 

[Dick]

You can rigorously show it by writing an explicit function for the full curve. But you really don't want to. What makes you think you can't pull the two copies of gamma3 apart? If it makes you feel better relabel one as gamma4. They are two separate curves, they aren't glued onto each other or anything.

 

[ehrenfest]

You can rigorously show it by writing an explicit function for the full curve. But you really don't want to. What makes you think you can't pull the two copies of gamma3 apart?

Because I have never seen any homotopy that moves apart two overlapping line segments. How do you homotope this into a circle:


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where you start on the left go to the right and come back.

EDIT: wait is it true that if the function restricted to each vertical and horizontal cross-section of the unit interval is continuous, then the function is continuous? Then it is clearly true. I will look for that theorem in Munkres.

 

[HallsofIvy]

Imagine cutting the line $\gamma_3$ into two parts so you have a kind of "bent" rectangle, then shrinking it down. Do you see how it fits onto $\gamma_5$?

 

[ehrenfest]

Imagine cutting the line $\gamma_3$ into two parts so you have a kind of "bent" rectangle, then shrinking it down. Do you see how it fits onto $\gamma_5$?

I don't really understand what you mean.

 

[ehrenfest]

AHA. Munkres Thereom 18.2 part f says that if I cross I can be written as the union of open sets such that the restriction of the homotopy to each open set is continuous, then the entire homopoty is continuous. So we just take as our open sets the part of I cross I that contains gamma_2 and a little bit of gamma_3, and then the part that contains a little bit of gamma_3, all of -gamma_1, and a littel bit of -gamma_3.

The restriction to each of those sets is clearly continuous.

I wish there were an easier way to prove that. Let me think...

 

[ehrenfest]

Alternatively, we can use the epsilon-delta formulation of continuity. I guess the main point is that continuity is a local property and that if you can put gamma3 and -gamma3 in disjoint balls, then the continuous movement of one is not related to the continuous movement of the other.

Say you have a point p fixed in I cross I and an epsilon greater than 0 and you want to find a delta-ball around around p in I cross I that gets mapped to a distance less than epsilon of H(p). It suffices to find a nbhd of p in I cross I where the restriction of the homotopy to that nbhd is continuous. That is what I was missing! And you can clearly do that for every point in I cross I for the example of this thread. The only points I was worried about were those on the preimage of \gamma_4. When you understand what I wrote in bold, then everything is crystal clear!

Please confirm that all of this is correct.

 

[Dick]

Maybe. But I think you are making this a little over elaborate. It's one single closed curve. There is nothing that prevents you from moving parts of the curve over each other. This is what I meant by saying 'it's not glued together'.