If anyone does this module please could you help me with this...
Define f: circle > circle to be the identity map i.e. f(z) = z
and g: circle > circle by g(z) = z^2
And let 1 be the base point for the circle (=s1).
Why is F(z,t) = z^(1+t) not a base point preserving homotopy from f to g??
Thanks

dave345
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 04042011 15:43

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 04042011 21:43
I may be wrong but perhaps you run into difficulties with continuity as t ranges from 0 to 1.

Totally Tom
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 07042011 20:16
(Original post by dave345)
If anyone does this module please could you help me with this...
Define f: circle > circle to be the identity map i.e. f(z) = z
and g: circle > circle by g(z) = z^2
And let 1 be the base point for the circle (=s1).
Why is F(z,t) = z^(1+t) not a base point preserving homotopy from f to g??
Thanks
(Original post by jdavi)
I may be wrong but perhaps you run into difficulties with continuity as t ranges from 0 to 1. 
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 09042011 14:56
Any tips for studying this module? All I can see is long proofs, few examples and useless past exams.

dave345
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 11042011 15:35
Ok thanks that makes sense, cheers.
Yeah it's a pain that the material for this course has changed so much this year, I've just been looking at the assignment sheets. Has anyone done much of sheet 4? I'm stuck on most of it, like how do you calculate the fundamental group of the torus with k points removed?? Does this use Van Kampen? 
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 11042011 16:10
(Original post by dave345)
Ok thanks that makes sense, cheers.
Yeah it's a pain that the material for this course has changed so much this year, I've just been looking at the assignment sheets. Has anyone done much of sheet 4? I'm stuck on most of it, like how do you calculate the fundamental group of the torus with k points removed?? Does this use Van Kampen?
Any ideas for Q3. on sheet 4? I suppose the identified edges are supposed to have the same letter, so that all edges are identified with one, if that makes sense. Can you just use the classification theorem? And for fundamental groups, do you use the same argument as for the projective plane and the torus (remove one point, then use van Kampen)?Last edited by rowzee; 11042011 at 16:17. 
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 11042011 17:15
Ok, I see why that works for one point but i don't quite see how you 'expand the holes' if there is more than one. As you say though it doesn't seem to be too formal.
For 3 I don't know how to show they aren't surfaces, i thought about the classification theorem but wasnt sure how to use it. Is it possble to show points on the boundry dont have neighbourhoods homeomorphic to a disc? Yeah thats what i did for the fundamental groups, I got C3 (cyclic group of order 3 for X1) and trivial group for X2. I guess as X1 doesn't have the same fundamental group as any of the surfaces in the classification thm it cant be a surface. 
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 11042011 17:41
(Original post by dave345)
Ok, I see why that works for one point but i don't quite see how you 'expand the holes' if there is more than one. As you say though it doesn't seem to be too formal.
For 3 I don't know how to show they aren't surfaces, i thought about the classification theorem but wasnt sure how to use it. Is it possble to show points on the boundry dont have neighbourhoods homeomorphic to a disc? Yeah thats what i did for the fundamental groups, I got C3 (cyclic group of order 3 for X1) and trivial group for X2. I guess as X1 doesn't have the same fundamental group as any of the surfaces in the classification thm it cant be a surface.
I will try to find the fundamental groups and compare answers. You're right about the locally homeomorphic condition, I think you can use it straight away on X2, since identifying two of the edges gives you the tip of a cone. Anyway, I think using the classification theorem once you know the fundamental groups should be fine (for X1 that is). 
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 11042011 18:51
Another way to think about this: the torus is , and the torus with a point removed is hence homotopy equivalent to for some interval (i.e. a cylinder). So it should be clear that . Removing successive points gives another point for loops to wind around in such a way that they cannot be shrunk to a point, i.e. adding (by direct sum) a factor of to the fundamental group.
Last edited by coffeym; 11042011 at 18:56. 
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 11042011 19:10
(Original post by coffeym)
Another way to think about this: the torus is , and the torus with a point removed is hence homotopy equivalent to for some interval (i.e. a cylinder). So it should be clear that . Removing successive points gives another point for loops to wind around in such a way that they cannot be shrunk to a point, i.e. adding (by direct sum) a factor of to the fundamental group.
EDIT: I don't think you're right at all, a punctured torus is homotopy equivalent to the figure 8 curve: if you take the identification space of the square and only consider the boundary of it (expand the hole to the whole interior) and glue it up, you clearly get a figure 8. Also, it would be the free product which is nonabelian, not the direct product.
Btw, I got and for the triangle spaces and as well, I just don't quite understand the step where you say what the induced loop is equal to.
Also, Q5. Sheet 4, how do you define the required homomorphism, which group do you take? I keep getting into trouble trying to send to the zero element in some free abelian group.Last edited by rowzee; 11042011 at 20:04. 
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 11042011 20:20
(Original post by rowzee)
Thanks, this is a much clearer way of thinking. Btw, what you wrote earlier and edited, was it wrong because the union was not disjoint? I believe that If you have a punctured cylinder, you can expand the hole so that you get 2 cylinders which meet at a single point and the whole thing is homotopy equivalent to the figure 8 curve. So you can keep puncturing the subcylinders and shrink them to always get a bouquet of circles
EDIT: I don't think you're right at all, a punctured torus is homotopy equivalent to the figure 8 curve: if you take the identification space of the square and only consider the boundary of it (expand the hole to the whole interior) and glue it up, you clearly get a figure 8. Also, it would be the free product which is nonabelian, not the direct product.
Btw, I got and for the triangle spaces and as well, I just don't quite understand the step where you say what the induced loop is equal to.
Also, Q5. Sheet 4, how do you define the required homomorphism, which group do you take? I keep getting into trouble trying to send to the zero element in some free abelian group. 
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 11042011 20:41
(Original post by coffeym)
My apologies, you are correct in terms of the figure 8 curve, I don't know what I was thinking! When you come to remove the second point, does it matter where you pick it from? If you choose the middle point on the figure 8, does this matter/make a difference? Also, what are you suggesting for the fundamental group in general? If you were to remove a point from one of the circles in the figure eight, then you get something homotopy equivalent to a circle, no?Last edited by rowzee; 11042011 at 20:43. 
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 11042011 20:48
(Original post by rowzee)
You can't remove points once you have retracted to the figure 8 curve, since you would be doing it posthomotopy equivalence and this screws thing up (I'm rather fuzzy about the reasons and this was not really taught). What you do is remove the k points at once in the identification space and when you're expanding the holes, make sure that you don't merge them. So if you have k points and expand them, you get a square that is subdivided into k horizontal (for simplicity) segments/holes (with k1 horizontal lines or boundaries of holes across). So when you do the 2 identifications, you get the figure 8 curve for the first hole +another circle for each each additional hole. So a bouquet of k+1 circles. (Hope this makes at least some sense). 
dave345
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 12042011 13:02
(Original post by rowzee)
Also, Q5. Sheet 4, how do you define the required homomorphism, which group do you take? I keep getting into trouble trying to send to the zero element in some free abelian group.
Did you manage to get anywhere with question 4? Think I can see what the group action should be, but how do you show the orbit sace is homeomorphic to the klein bottle? 
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 12042011 16:33
(Original post by dave345)
Did you manage to get anywhere with question 4? Think I can see what the group action should be, but how do you show the orbit sace is homeomorphic to the klein bottle?
Edit: looks good for Q5, thanks!Last edited by rowzee; 12042011 at 16:53. 
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 12042011 23:58
(Original post by rowzee)
I have no idea. The examples in the notes (e.g. ) use explicit maps and the Topologist's First Isomorphism Theorem, so I'm guessing something along those lines. Of course, you can say right away that since the action is free and properly discontinuous, but you can't conclude isomorphic fundamental groups=>isomorphic spaces.
Edit: looks good for Q5, thanks! 
jimbob_barnes
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 22042011 11:09
God this module is such bull****. I have looked at every past exam paper and they are sooooooooooo much easier than the assignment sheets. Not that the assignment sheets are any good either as there are stupid questions like "what has this got to do with the quaternions".
I have issues with Sheet 4 Q2 part 1. The notes say that S^n\{a,b} has trivial fundamental group. But I think this is wrong for n=2 as in this case S^n\{a} is homeomorphic to R^2 and thus S^n\{a,b} is homomorphic to the punctured plane, which has fundamental group Z. Then as you remove more and more points you get more punctures so you get something homotopy equivalent to a onepoint sum of circles. Then S^n\{k_1,...,k_n} has trivial fundamental group for k=0,1 and for k>1 it is the free product of k copies of Z.
If n>2 I see why the fundamental group is trivial so that is all ok.
I think I'm gonna beg my tutor for a revision session, hopefully he can help. 
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 22042011 14:00
(Original post by jimbob_barnes)
I have issues with Sheet 4 Q2 part 1. The notes say that S^n\{a,b} has trivial fundamental group. But I think this is wrong for n=2 as in this case S^n\{a} is homeomorphic to R^2 and thus S^n\{a,b} is homomorphic to the punctured plane, which has fundamental group Z. Then as you remove more and more points you get more punctures so you get something homotopy equivalent to a onepoint sum of circles. Then S^n\{k_1,...,k_n} has trivial fundamental group for k=0,1 and for k>1 it is the free product of k copies of Z.
If n>2 I see why the fundamental group is trivial so that is all ok. 
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 23042011 17:59
(Original post by jimbob_barnes)
God this module is such bull****. I have looked at every past exam paper and they are sooooooooooo much easier than the assignment sheets. Not that the assignment sheets are any good either as there are stupid questions like "what has this got to do with the quaternions".
I have issues with Sheet 4 Q2 part 1. The notes say that S^n\{a,b} has trivial fundamental group. But I think this is wrong for n=2 as in this case S^n\{a} is homeomorphic to R^2 and thus S^n\{a,b} is homomorphic to the punctured plane, which has fundamental group Z. Then as you remove more and more points you get more punctures so you get something homotopy equivalent to a onepoint sum of circles. Then S^n\{k_1,...,k_n} has trivial fundamental group for k=0,1 and for k>1 it is the free product of k copies of Z.
If n>2 I see why the fundamental group is trivial so that is all ok.
I think I'm gonna beg my tutor for a revision session, hopefully he can help. 
dave345
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 25042011 16:56
Has anyone got anywhere at all with quesion 4, sheet 4? Soooo hard!
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