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Thread: Algebraic topology (an introduction)

  1. #1 Algebraic topology (an introduction) 
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    Warning: This thread may be pretty heavy going for some; please ignore it if you don’t feel up to it.

    Let be defined by for each . Then as t varies continuously from 0 to 1, p(t) moves continuously from (1,0) to (−1,0) along the upper semicircle of the unit circle centred at the origin. We say that p is a path in from (1,0) to (−1,0). In general:

    Definition 1: If a and b are points in a topological space X, a path in X from a to b is a continuous function such that p(0) = a and p(1) = b. If a = b, the path p is called a loop at a.

    Properties of paths:

    (i) If p is a path as defined above, the inverse path of p is the path in X from b to a defined by for each .

    (ii) Let p be a path from a to b and q be a path from b to c. Note that p(1) = b = q(0). The product path of p and q is the path from a to c defined as follows:



    The fact that p*q is continuous is due to a result in general topology sometimes called the “pasting lemma”ť:

    Lemma 1: Let A and B be closed subsets of a topological space X such that , and suppose and are continuous functions to a topological space Y such that f(x) = g(x) for all xAB. Then the function where h(x) = f(x) if xA and h(x) = g(x) if xB is continuous.

    (iii) The constant loop mapping all points in [0,1] to a is called the null path at a.

    Exercise 1: Let p be a path from a to b, q be a path from b to c, and r be a path from c to d in X. If or , explain why is NOT equal to .


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  3. #2  
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    Now let’s return to the example at the beginning of the previous post: the path . If we set q(t) = −p(t) for each t ∊ [0,1], we have another path from (1,0) to (−1,0) in <sup>2</sup>, this one the lower semicircle of the same unit circle.

    And now, for each s ∊ [0,1], let’s define a mapping by for all t ∊ [0,1]. Notice that f<sub>0</sub>(t) = p(t) and f<sub>1</sub>(t) = q(t) for all t ∊ [0,1]. So f<sub>0</sub> is the path p and f<sub>1</sub> is the path q, and for each s ∊ [0,1] f<sub>s</sub> is a path tracing out a semiellipse. (When s = ˝, the semiellipse is degenerate, being the straight line from (1,0) to (−1,0).)

    What happens is that as s varies continously from 0 to 1, the path p is continously deformed into q via the various intermediate paths f<sub>s</sub>.

    In general, continuous functions between topological spaces (not just paths) can be continuously deformed into one other in this way (if certain conditions hold). This is the basis of homotopy theory.

    Definition 2: Let X and Y be topological spaces and continuous functions. Then f is said to be homotopic to g (written ) iff

    (i) given any s ∊ [0,1], there exists a continuous function with f<sub>0</sub> = f and f<sub>1</sub> = g, and

    (ii) there exists a continuous function such that F(x,s) = f<sub>s</sub>(x) for each xX, s ∊ [0,1].

    The function F is called a homotopy from f to g.

    Notes:

    (i) The existence of intermediate continuous functions f<sub>s</sub> for all s ∊ [0,1] does not mean that fg – the “deformation”ť process still has to be carried out smoothly. This is what the continuous function F ensures.

    (ii) If X is homeomorphic to Y and each intermediate f<sub>s</sub> is a homeomorphism from X to Y, f is said to be isotopic to g, and the function F is called an isotopy from f to g.

    (iii) The class of all continuous functions from X to Y is denoted Hom(X,Y). This class can be regarded as a category (see Guitarist’s excellet thread on category theory) with continuous functions as objects and homotopies as morphisms; the isomorphisms of this category are then the isotopies of Hom(X,Y). :-D

    Exercise 2: Prove that the relation “is homotopic to”ť is an equivalence relation on Hom(X,Y). (You will need to use the “pasting lemma”ť (Lemma 1) to prove transitivity.)


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  4. #3 Re: Algebraic topology 
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    Thanks for posting this. I followed it this far:

    Quote Originally Posted by JaneBennet

    (ii) Let p be a path from a to b and q be a path from b to c. Note that p(1) = b = q(0). The product path of p and q is the path p*q from a to c defined as follows:



    The fact that p*q is continuous is due to a result in general topology sometimes called the “pasting lemma”:
    Can you elaborate on the definition in LaTeX and what it means? t looks like a new operation to me. I understand that p(1)=q(0), and that the two paths combine to make a new path, but not the LaTeX expression.

    Thanks
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  5. #4  
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    I made a slight mistake with my equation, silly me.

    I have fixed it now; the LHS should be p*q(t). ( is a function .)
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  6. #5  
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    So basically the notation for the product path is just listing the 'component' paths p(t) and q(t), making up the overall path p*q(t)?

    Why not just write the upper path, p, as p(t) 0≤t≤1, and the lower path, q, as q(t) 0≤t≤1 ?

    What if there were a third path s(t) from c to d. would you write it as:

    p*q*s(t) =

    p(t) 0≤t≤1
    q(t) 0≤t≤1
    s(t) 0≤t≤1


    (but with the curly bracket)?
    Chance favours the prepared mind.
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  7. #6  
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    Quote Originally Posted by bit4bit
    Why not just write the upper path, p, as p(t) 0≤t≤1, and the lower path, q, as q(t) 0≤t≤1 ?
    Well, no. You can only do paths p and q separately. If you do them together, the function p*q won’t be well defined. :? p*q means: do path p first, then q.

    You see, t is a variable that goes continuously from 0 to 1. As it does so, the path p traces out a continuous “route”ť in the topological space X from a to b. Similarly the path q traces out a continuous “route”ť in X from b to c.

    Now suppose you want to get from a to c. Since the path p ends at b and the path q starts at b, you can make your journey via b using paths p and q. But there’s a catch: you still have to make your journey as t moves from 0 to 1 – no faster, no slower. So, what do you do? Well, the trick is to move twice as fast as you would if you were just taking path p or path q alone! :P

    So if you go through path p twice as fast, you’ll reach b at t = ½ rather than t = 1. This will leave you time to go through path q, moving at the same double speed, and get to c when t hits 1. This is why you have p(2t) rather than p(t) in the formula for the product path – because you want to move at double speed. Same for q – except that for q, you start at t = ½ rather than 0, so the formula is q(2(t−½)) = q(2t−1) instead.

    Similarly if you have three paths. But note Exercise 1: the product of paths is not associative!
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  8. #7  
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    Jane: Good thread you have going here. Homotopy theory is a lot of fun.

    Keep it coming!
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  9. #8  
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    Yeah, I intend to develop the thread at least up to the definition of fundamental group. (Maybe further if I feel up to it.)

    Anyway, I’ve reviewed the material that I need and found that there is one very important result I need which I have to mention now. The proof is anything but straightforward. :?

    Theorem 1: Let p, q and r be paths as described in Exercise 1. Then .

    Proof: From the definition of product path, we have the following:





    So you can see why and are not equal in general: with the former path, you go twice as quickly from a to b and twice as slowly from c to d as with the latter. There’s the answer to Exercise 1. (Did you get it?)

    The two paths are nevertheless homotopic. So what might a homotopy be? Well, I have the answer in my book (A First Course in Algebraic Topology by B.K. Lahiri) so I can tell you what it is:



    Goodness me! I’d never have come up with that one myself in a million years.
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  10. #9  
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    Help!!! Jane your photobucket images don't render for me. Wht?
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  11. #10  
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    It’s working fine for me.
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  12. #11  
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    Quote Originally Posted by JaneBennet
    Well, no. You can only do paths p and q separately. If you do them together, the function p*q won’t be well defined. :? p*q means: do path p first, then q.

    You see, t is a variable that goes continuously from 0 to 1. As it does so, the path p traces out a continuous “route” in the topological space X from a to b. Similarly the path q traces out a continuous “route” in X from b to c.

    Now suppose you want to get from a to c. Since the path p ends at b and the path q starts at b, you can make your journey via b using paths p and q. But there’s a catch: you still have to make your journey as t moves from 0 to 1 – no faster, no slower. So, what do you do? Well, the trick is to move twice as fast as you would if you were just taking path p or path q alone! :P

    So if you go through path p twice as fast, you’ll reach b at t = ½ rather than t = 1. This will leave you time to go through path q, moving at the same double speed, and get to c when t hits 1. This is why you have p(2t) rather than p(t) in the formula for the product path – because you want to move at double speed. Same for q – except that for q, you start at t = ½ rather than 0, so the formula is q(2(t−½)) = q(2t−1) instead.

    Similarly if you have three paths. But note Exercise 1: the product of paths is not associative!
    Thanks Jane, I got it now.
    Chance favours the prepared mind.
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  13. #12  
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    You’re welcome, bit4bit. 8)

    Now let’s explore some properties of null paths. Recall that a null path at aX is the constant path where for all t ∊ [0,1].

    Theorem 2: For any paths p in X, and .

    Proof: Take where



    Then F is a homotopy from p*ε<sub>p(1)</sub> to p.

    Similarly, where



    is a homotopy from ε<sub>p(0)</sub>*p to p.

    Theorem 3: and .

    Proof: Note that



    Now, what trick is it this time? Well, I can tell you:



    That’s the homotopy that does the job.

    To prove the other case, replace p by in the above, noting that and . Hmm, this will do for the next exercise.

    Exercise 3: (i) Show that for any path p in X. (ii) Prove that if p and q are homotopic paths, then their inverse paths are homotopic.
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  14. #13  
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    Quote Originally Posted by Guitarist
    Help!!! Jane your photobucket images don't render for me. Wht?
    It shouldn’t be a problem now. I’ve edited my posts and rewritten much of my stuff in TeX mode!

    Next topic: Relative homotopy and equivalent paths. 8)
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  15. #14  
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    Definition 2: Let and . If for all , then is called a homotopy from f to g relative to A iff H is a homotopy from f to g such that for all .

    If f is homotopic to g relative to A, we’ll write .

    In particular, if are paths such that and , then p are q are said to be equivalent iff they are homotopic relative to , i.e. iff .

    We come now to something very important.

    Let X be a topological space and a be a fixed point in X. Consider the set of all paths starting and ending at a (i.e. all loops at a). Define a relation ~ on this set by p and q are equivalent. Then ~ is an equivalence relation (cf Exercise 2); write to for the equivalence class containing the path p, and denote the set of all equivalence classes by .

    Now we define a binary operation * on as follows:



    In other words, the “product”ť of the equivalence classes of two paths shall be the equivalence class of the product path. The very first question we have to ask is: is this operation well defined? In other words, will we get the same result for the operation even if we take different paths as “representatives”ť for the same equivalence classes?

    And the answer this is … yes! :-D The operation is well defined thanks to the following:

    Theorem 4: Let p and q be equivalent paths from a to b in the topological space X (so and ). Similarly, let r and s be equivalent paths from b to c in X. Then .
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    Proof of Theorem 4: Let be homotopies relative to {0,1} between p and q and between r and s respectively. Thus, for all , we have the following:











    Now let be defined as follows:



    Note that so H is well defined and continuous at .

    Now let’s check that we have everything else.














    Hence H is a homotopy from to . Let’s verify that it’s a homotopy relative to {0,1}.





    And so it is. The theorem is proved. 8)

    Very important note:
    If p and q (likewise r and s) are only homotopic, not equivalent, the above proof will not work. This is because H may then not be continuous (or even well defined) at . Indeed, it is false that : the homotopies must be relative to {0,1}. Here is an example to show why.

    Let and let be as we’ve defined in the opening post of this thread: . Now let , , . Then p and q are homotopic, but not equivalent; similarly for r and s. Now is the path tracing out the unit circle centred at (0,0) while traces out the unit circle centred at (0,2). It is clear that . The former “encloses”ť the point (0,0), which is not in X, so there is no way it can be continuously deformed into a path that doesn’t “enclose”ť (0,0).
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  17. #16  
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    Right, so the binary operation * we defined on by



    is well defined. Then we have …

    Theorem 5: is a group.

    Proof:

    By Theorem 1, * is associative.

    By Theorem 2, is the identity element.

    By Theorem 3, the inverse of is .

    is called the fundamental group of X at a.

    Various interesting and important results can be derived from the fundamental group: e.g. every continuous function induces a homomorphism defined by (where denotes functional composition), and if f is homeomorphism, then is an isomorphism. It follows that one can show that two topological spaces are not homeomorphic by showing that corresponding fundamental groups are not isomorphic! Isn’t that marvellous?

    Alas, showing that two topological spaces are homeomorphic may not be that easy: it is possible for two fundamental groups to be isomorphic while the original topological spaces are not homeomorphic. At least we have achieved the first step towards the motivating goal of algebraic topology, namely using tools from abstract algebra to solve problems in general topology. 8)
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  18. #17  
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    Nice work, Jane. Let me add a couple of fun points, but in doing so I will need to change the notation slightly, just for clarity.

    It is usual to define the fundamental group with reference to a "privileged" point in X, called the base-point. That is, the group is the fundamental group at the point .

    Now suppose that is some other point in X Then, provided only that is path-connected, I may define the path

    So now suppose that is an homotopy class of loops at .

    Then I will have an induced map given by .

    That is, go from to using , make the loop using the path , and than back to using .

    In other words, this defines a loop based at .

    It is simps to show that this defines a group homomorphism with inverse.

    Mmm, run out of energy for my second "fun point"

    Later
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  19. #18  
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    W00t! Thanks for the fun point, Guitarist!

    Yeah, I was using rather than to denote the base point of my fundamental group.

    While you catch your breath for your second fun point, here’s a fun point of mine regarding path-connectedness:

    Fun Theorem: Every topological space is the disjoint union of path-connected subspaces.

    Proof: Define the relation ~ on a topological space X by there is a path from a to b in X. Then ~ is an equivalence relation and the equivalence classes are path-connected subspaces of X.
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  20. #19  
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    Jane; This nice, though slightly counter-intuitive. It is simply to state the fact, that we all know, that an equivalence relation on any set is a partition, under which operation, partitions are disjoint, by definition.

    Now let me gird my loins (as it were)

    Suppose are two pointed topological spaces, where as before are referred to as "base-points".

    Suppose now that we have a continuous base-point-respecting map .

    Then I will have the group homomorphism

    Then, for any loop I will have that is a loop in , where .

    Define this as the map , given by .

    Then, if it is the case that and , then .

    Moreover, looking at the constant loop , I will see that .

    All this is sufficient to define as functor from the category of pointed spaces to the category of groups.

    In fact, historically, this was the first description of such a beast, though Poincaré never used the term
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  21. #20  
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    Quote Originally Posted by JaneBennet
    Fun Theorem: Every topological space is the disjoint union of path-connected subspaces.

    Proof: Define the relation ~ on a topological space X by there is a path from a to b in X. Then ~ is an equivalence relation and the equivalence classes are path-connected subspaces of X.
    However, each one of these path connected components may just be a singleton (totally disconnected spaces etc...). Its a case many people forget about
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  22. #21  
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    @river_rat: Yeah, with different topological spaces you get an interesting selection of path-connected subspaces, ranging from singleton subspaces for totally disconnected spaces to just {X} for a path-connected space X.

    @Guitarist: Nice one.

    Anyway …

    The pasting lemma (Lemma 1) has been used quite a lot in the development of homotopy theory. Since it’s so important, it might be worth giving a proof of it. This I’ll do now; first I’ll re-state it.

    Lemma 1: Let A and B be closed subsets of a topological space X such that , and suppose and are continuous functions to a topological space Y such that f(x) = g(x) for all xAB. Then the function where h(x) = f(x) if xA and h(x) = g(x) if xB is continuous.

    Proof:

    For this proof, we use this result: is continuous if and only if for any closed subset , is closed in . (Recall that denotes the set .)

    Let be closed in Y and consider . Then is closed in ; where is closed in . Hence is closed in . Similarly (where is closed in ) is closed in . Thus is closed in . QED.

    Exercise 4: In the proof of the above theorem, show that .

    Exercise 5: Let h be any function from any set X to any set Y.

    (i) Show that if , we have . (In complement notation: .)

    (ii) Hence show that if X and Y are topological spaces, h is a continuous function if and only if for any closed subset , is closed in .
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  23. #22  
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    Right. Let’s take a closer look at some of the fun points that have been presented.

    Guitarist’s first fun point says that if two points in a topological space can be connected by a path, then the fundamental groups at these points are isomorphic. Combining this with my own fun point – namely every topological space can be partitioned into path-connected subspaces – we see that we can analyse all the fundamental groups of a topological space just by analysing the fundamental groups of distinct path-connected subspaces.

    For the next fun point, Guitarist pointed out the induced homomorphism given topological spaces X and Y, the point and the continuous function . The induced homorphism is defined by for each loop at a ( denotes composition of functions; the composite of two continuous functions is continuous). It is a homomorphism because for any loops p and q at a, (as can be easily checked).

    Now suppose f is a homeomorphism. Suppose . Then if F is a homotopy from to relative to {0,1}, is a homotopy from to relative to {0,1}; hence . And if r is a loop at , is a loop at a and (so ). Hence, if f is a homeomorphism, the induced homomorphism is an isomorphism. 8)
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