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    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
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    I may be wrong but perhaps you run into difficulties with continuity as t ranges from 0 to 1.
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    (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
    So the issue is the end points. I'm pretty sure that if you've got a homotopy from f to g the endpoints of the homotopy function have to be fixed. In your example one of them slides along the circle and back to the base point.
    (Original post by jdavi)
    I may be wrong but perhaps you run into difficulties with continuity as t ranges from 0 to 1.
    The exponential map is continuous so this is not the issue.
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    Any tips for studying this module? All I can see is long proofs, few examples and useless past exams.
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    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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    (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?
    There appears to be very little formality in these kind of arguments. Basically, if you remove 1 point and expand the hole, you get something homotopy equivalent to figure 8, so the fundamental group is the free product Z*Z. As you keep removing points, you increase the number of 'petals' (i.e. loops starting at the centre of the figure 8). So with k removed points, the fundamental group is the free product Z*Z*...*Z k+1 times.

    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)?
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    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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    (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 think searching 'torus with points removed fundamental group' on the internet will give a better answer than I can.

    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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    Another way to think about this: the torus is S^1 \times S^1 , and the torus with a point removed is hence homotopy equivalent to S^1 \times I for some interval I (i.e. a cylinder). So it should be clear that  \pi_1 (S^1 \times I) \cong \mathbb{Z} . 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 \mathbb{Z} to the fundamental group.
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    (Original post by coffeym)
    Another way to think about this: the torus is S^1 \times S^1 , and the torus with a point removed is hence homotopy equivalent to S^1 \times I for some interval I (i.e. a cylinder). So it should be clear that  \pi_1 (S^1 \times I) \cong \mathbb{Z} . 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 \mathbb{Z} to the fundamental group.
    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 sub-cylinders 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 non-abelian, not the direct product.

    Btw, I got C_3 and \{1\} for the triangle spaces X_1 and X_2 as well, I just don't quite understand the step where you say what the induced loop i_*([c]) 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 a_1a_2...a_na_na_{n-1}...a_1 to the zero element in some free abelian group.
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    (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 sub-cylinders 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 non-abelian, not the direct product.

    Btw, I got C_3 and \{1\} for the triangle spaces X_1 and X_2 as well, I just don't quite understand the step where you say what the induced loop i_*([c]) 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 a_1a_2...a_na_na_{n-1}...a_1 to the zero element in some free abelian group.
    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?
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    (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?
    You can't remove points once you have retracted to the figure 8 curve, since you would be doing it post-homotopy 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 k-1 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).
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    (Original post by rowzee)
    You can't remove points once you have retracted to the figure 8 curve, since you would be doing it post-homotopy 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 k-1 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).
    Yes, this is a nice explanation. For some reason I seem to have forgotten much of what I learnt about homotopy equivalence. But then, last year's course didn't have much reason to worry about it except that homotopy equivalent spaces have isomorphic fundamental groups.
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    (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 a_1a_2...a_na_na_{n-1}...a_1 to the zero element in some free abelian group.
    Can you take diect sum of n lots of Z/2Z ?

    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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    (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?
    I have no idea. The examples in the notes (e.g. \textbf{R}/\textbf{Z}\cong \textbf{S}^1) 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 \pi_1(\textbf{R}^2/G)\cong G=\pi_1(\textbf{K}) since the action is free and properly discontinuous, but you can't conclude isomorphic fundamental groups=>isomorphic spaces.

    Edit: \oplus_{i=1}^{n}\textbf{Z}/2 \textbf{Z} looks good for Q5, thanks!
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    (Original post by rowzee)
    I have no idea. The examples in the notes (e.g. \textbf{R}/\textbf{Z}\cong \textbf{S}^1) 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 \pi_1(\textbf{R}^2/G)\cong G=\pi_1(\textbf{K}) since the action is free and properly discontinuous, but you can't conclude isomorphic fundamental groups=>isomorphic spaces.

    Edit: \oplus_{i=1}^{n}\textbf{Z}/2 \textbf{Z} looks good for Q5, thanks!
    Yeah thats exactly what I was thinking, dont really know how you come up with a map into the klein bottle though?!
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    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 one-point 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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    (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 one-point 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 agree with all of that.
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    (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 one-point 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.
    You've misread the notes, I believe. That should say the fundamental group: pi[S^n\{a}, b] meaning the fundamental group of the n-sphere with point a removed with base-point b, where b=/=a. Hopefully this makes the homework question make sense.
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    Has anyone got anywhere at all with quesion 4, sheet 4? Soooo hard!
 
 
 
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