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JaneBennet

Posted: Fri Jul 04, 2008 6:19 am Post subject: Algebraic topology

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Joined: 06 Apr 2008
Posts: 792

Warning: This thread may be pretty heavy going for some; please ignore it if you don’t feel up to it. Smile

Let $p:[0,1]\to\mathbb{R}^2$ be defined by $p(t)=(\cos{\pi t},\sin{\pi t})$ for each $t\in[0,1]$ . 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 $\mathbb{R}^2$ 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 $p:[0,1]\to X$ 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 $\overline{p}$ in X from b to a defined by $\overline{p}(t)=p(1-t)$ for each $t\in[0,1]$ .

(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 $p\star q$ from a to c defined as follows:

$p\star q\ =\ \left\{\begin{array}{ll}p(2t) & 0\leq t\leq\frac{1}{2}\\\\q(2t-1) & \frac{1}{2}\leq r\leq0\end{array}$

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

Lemma 1: Let A and B be closed subsets of a topological space X such that $X=A\cup B$ , and suppose $f:A\to Y$ and $g:B\to Y$ are continuous functions to a topological space Y such that f(x) = g(x) for all x ∊ A ∩ B. Then the function $h:X\to Y$ where h(x) = f(x) if x ∊ A and h(x) = g(x) if x ∊ B 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 $a\neq b$ or $c\neq d$ , explain why $(p\star q)\star r$ is NOT equal to $p\star (q\star r)$ .
_________________

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Did You Know?
Fact of the day: Homotopy

Last edited by JaneBennet on Thu Aug 07, 2008 3:49 am; edited 4 times in total

JaneBennet

Posted: Fri Jul 04, 2008 8:23 am Post subject:

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Now let’s return to the example at the beginning of the previous post: the path $p(t)=\left(\cos{\pi t},\sin{\pi t}\right)$ . If we set q(t) = −p(t) for each t ∊ [0,1], we have another path from (1,0) to (−1,0) in ℝ², this one the lower semicircle of the same unit circle.

And now, for each s ∊ [0,1], let’s define a mapping $f_s:[0,1]\to\mathbb{R}^2$ by $f(s,t)=\left(\cos{\pi t},(1-2s)\sin{\pi t}\right)$ for all t ∊ [0,1]. Notice that f₀(t) = p(t) and f₁(t) = q(t) for all t ∊ [0,1]. So f₀ is the path p and f₁ is the path q, and for each s ∊ [0,1] f_s is a path tracing out a semiellipse. (When s = ½, 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_s.

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 $f,g:X\to Y$ continuous functions. Then f is said to be homotopic to g (written $f\simeq g$ ) iff

(i) given any s ∊ [0,1], there exists a continuous function $f_s:X\to Y$ with f₀ = f and f₁ = g, and

(ii) there exists a continuous function $F:X\times[0,1]\to Y$ such that F(x,s) = f_s(x) for each x ∊ X, s ∊ [0,1].

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

Notes:

(i) The existence of intermediate continuous functions f_s for all s ∊ [0,1] does not mean that f ≃ g – the “deformation” 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_s 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). Very Happy

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

A problem worthy of attack
Proves its worth by fighting back.
(Piet Hein)

Did You Know?
Fact of the day: Homotopy

Last edited by JaneBennet on Thu Aug 07, 2008 4:05 am; edited 4 times in total

bit4bit

Posted: Fri Jul 04, 2008 9:34 am Post subject: Re: Algebraic topology

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Joined: 14 Jul 2007
Posts: 623

Thanks for posting this. I followed it this far:

JaneBennet wrote:

(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 â€œpasting lemmaâ€:

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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JaneBennet

Posted: Fri Jul 04, 2008 9:47 am Post subject:

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Posts: 792

I made a slight mistake with my equation, silly me. Embarassed

I have fixed it now; the LHS should be p*q(t). Smile

( $p\star q$ is a function $p,q:[0,1]\to X$ .)
_________________

A problem worthy of attack
Proves its worth by fighting back.
(Piet Hein)

Did You Know?
Fact of the day: Homotopy

Last edited by JaneBennet on Thu Aug 07, 2008 3:52 am; edited 2 times in total

bit4bit

Posted: Fri Jul 04, 2008 10:13 am Post subject:

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Joined: 14 Jul 2007
Posts: 623

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.

JaneBennet

Posted: Fri Jul 04, 2008 11:02 am Post subject:

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Joined: 06 Apr 2008
Posts: 792

bit4bit wrote:

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. Confused

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! Razz

So if you go through path p twice as fast, you’ll reach b at t = Â½ 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 = Â½ rather than 0, so the formula is q(2(t−Â½)) = q(2t−1) instead. Wink

Similarly if you have three paths. But note Exercise 1: the product of paths is not associative! Shocked

_________________

A problem worthy of attack
Proves its worth by fighting back.
(Piet Hein)

Did You Know?
Fact of the day: Homotopy

Last edited by JaneBennet on Thu Aug 07, 2008 4:07 am; edited 2 times in total

Guitarist

Posted: Fri Jul 04, 2008 12:17 pm Post subject:

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Joined: 08 Jun 2005
Posts: 716

Jane: Good thread you have going here. Homotopy theory is a lot of fun.

Keep it coming!

JaneBennet

Posted: Fri Jul 04, 2008 5:56 pm Post subject:

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Joined: 06 Apr 2008
Posts: 792

Yeah, I intend to develop the thread at least up to the definition of fundamental group. (Maybe further if I feel up to it.) Smile

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. Confused

Theorem 1: Let p, q and r be paths as described in Exercise 1. Then $(p\star q)\star r\simeq p\star(q\star r)$ .

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

$(p\star q)\star r\<img src="images/smiles/icon_sad.gif" alt="Sad" border="0" />t)\ =\ \left\{\begin{array}{ll}p(4t) & 0\leq t\leq\frac{1}{4}\\\\ q(4t-1) & \frac{1}{4}\leq t\leq\frac{1}{2}\\\\ r(2t-1) & \frac{1}{2}\leq t\leq1\end{array}$

$p\star(q\star r)\<img src="images/smiles/icon_sad.gif" alt="Sad" border="0" />t)\ =\ \left\{\begin{array}{ll}p(2t) & 0\leq t\leq\frac{1}{2}\\\\ q(4t-2) & \frac{1}{2}\leq t\leq\frac{3}{4}\\\\ r(4t-3) & \frac{3}{4}\leq t\leq1\end{array}$

So you can see why $(p\star q)\star r$ and $p\star(q\star r)$ 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?) Very Happy

The two paths are nevertheless homotopic. So what might a homotopy $F:[0,1]\times[0,1]\to X$ 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:

$F(s,t)\ =\ \left\{\begin{array}{ll}p\left(\frac{4s}{1+t}\right) & 0\leq s\leq\frac{1+t}{4}\\\\ q\left(4s-t-1\right) & \frac{1+t}{4}\leq s\leq\frac{2+t}{4}\\\\ r\left(\frac{4s-t-2}{2-t}\right) & \frac{2+t}{4}\leq s\leq1\end{array}$

Goodness me! I’d never have come up with that one myself in a million years. Shocked

Guitarist

Posted: Sat Jul 05, 2008 3:01 am Post subject:

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Joined: 08 Jun 2005
Posts: 716

Help!!! Jane your photobucket images don't render for me. Wht?

JaneBennet

Posted: Sat Jul 05, 2008 3:39 am Post subject:

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Joined: 06 Apr 2008
Posts: 792

It’s working fine for me. Surprised

bit4bit

Posted: Sat Jul 05, 2008 5:57 am Post subject:

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Joined: 14 Jul 2007
Posts: 623

JaneBennet wrote:

Well, no. You can only do paths p and q separately. If you do them together, the function p*q wonâ€™t be well defined. Confused

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 â€œrouteâ€ in the topological space X from a to b. Similarly the path q traces out a continuous â€œrouteâ€ 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 thereâ€™s a catch: you still have to make your journey as t moves from 0 to 1 â€“ 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! Razz

So if you go through path p twice as fast, youâ€™ll reach b at t = Â½ 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 â€“ because you want to move at double speed. Same for q â€“ except that for q, you start at t = Â½ rather than 0, so the formula is q(2(t−Â½)) = q(2t−1) instead. Wink

Similarly if you have three paths. But note Exercise 1: the product of paths is not associative! Shocked

Thanks Jane, I got it now.
_________________
Chance favours the prepared mind.

JaneBennet

Posted: Sat Jul 05, 2008 11:50 am Post subject:

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Joined: 06 Apr 2008
Posts: 792

You’re welcome, bit4bit. Cool

Now let’s explore some properties of null paths. Recall that a null path at a ∊ X is the constant path $\epsilon_a:[0,1]\to X$ where $\epsilon_a(t)=a$ for all t ∊ [0,1].

Theorem 2: For any paths p in X, $p\star\epsilon_{p(1)}\simeq p$ and $\epsilon_{p(0)}\star p\simeq p$ .

Proof: Take $F:[0,1]\times[0,1]\to X$ where

$F(s,t)\ =\ \left\{\begin{array}{ll}p\left(\frac{2s}{1+t}\right) & 0\leq s\leq\frac{(1+t)}{2}\\\\ p(1) & \frac{1+t}{2}\leq s\leq1\end{array}$

Then F is a homotopy from p*ε_p(1) to p.

Similarly, $G:[0,1]\times[0,1]\to X$ where

$G(s,t)\ =\ \left\{\begin{array}{ll}p(0) & 0\leq s\leq\frac{1-t}{2}\\\\ p\left(\frac{2s+t-1}{1+t}\right) & \frac{1+t}{2}\leq s\leq1\end{array}$

is a homotopy from ε_p(0)*p to p.

Theorem 3: $p\star\overline{p}\simeq\epsilon_{p(0)}$ and $\overline{p}\star p\simeq\epsilon_{p(1)}$ .

Proof: Note that

$p\star\overline{p}\<img src="images/smiles/icon_sad.gif" alt="Sad" border="0" />t)\ =\ \left\{\begin{array}{ll}p(2t) & 0\leq t\leq\frac{1}{2}\\\\ p(2-2t) & \frac{1}{2}\leq t\leq1\end{array}$

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

$F(s,t)\ =\ \left\{\begin{array}{ll}p\left(2s(1-t)\right) & 0\leq s\leq\frac{1}{2}\\\\ p\left((2-2s)(1-t)\right) & \frac{1}{2}\leq s\leq1\end{array}$

That’s the homotopy that does the job.

To prove the other case, replace p by $\overline{p}$ in the above, noting that $\overline{p}(0)=p(1)$ and $\overline{\overline{p}}=p$ . Hmm, this will do for the next exercise. Very Happy

Exercise 3: (i) Show that $\overline{\overline{p}}=p$ for any path p in X. (ii) Prove that if p and q are homotopic paths, then their inverse paths are homotopic.
_________________

A problem worthy of attack
Proves its worth by fighting back.
(Piet Hein)

Did You Know?
Fact of the day: Homotopy

Last edited by JaneBennet on Thu Aug 07, 2008 4:09 am; edited 1 time in total

JaneBennet

Posted: Sun Jul 06, 2008 4:04 pm Post subject:

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Joined: 06 Apr 2008
Posts: 792

Guitarist wrote:

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. Cool

JaneBennet

Posted: Sat Jul 12, 2008 12:36 pm Post subject:

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Joined: 06 Apr 2008
Posts: 792

Definition 2: Let $f,g:X\to Y$ and $A\subseteq X$ . If f(x)=g(x)

for all $x\in A$ , then $H:X\times[0,1]\to Y$ is called a homotopy from f to g relative to A iff H is a homotopy from f to g such that H(x,t)=f(x)=g(x)

for all $x\in A,\ t\in[0,1]$ .

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

In particular, if $p,q:[0,1]\to X$ are paths such that p(0)=q(0)

and

, then p are q are said to be equivalent iff they are homotopic relative to $\{0,1\}$ , i.e. iff $p\simeq_{\{0,1\}}q$ .

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 $p\sim q\ \Leftrightarrow$ p and q are equivalent. Then ~ is an equivalence relation (cf Exercise 2); write [p]

to for the equivalence class containing the path p, and denote the set of all equivalence classes by $\pi_1(X,a)$ .

Now we define a binary operation * on $\pi_1(X,a)$ as follows:

$\forall\,[p_1],[p_2]\in\pi_1(X,a),\ [p_1]\ast[p_2]\ =\ [p_1\star p_2]$

In other words, the “product” 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” for the same equivalence classes?

And the answer this is … yes! Very Happy

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 p(0)=q(0)=a

and

). Similarly, let r and s be equivalent paths from b to c in X. Then $p\simeq_{\{0,1\}}q\ \mbox{and}\ r\simeq_{\{0,1\}}s\ \Rightarrow\ p\star r\simeq_{\{0,1\}}q\star s$ .
_________________

A problem worthy of attack
Proves its worth by fighting back.
(Piet Hein)

Did You Know?
Fact of the day: Homotopy

Last edited by JaneBennet on Thu Aug 07, 2008 4:11 am; edited 2 times in total

JaneBennet

Posted: Sat Jul 12, 2008 2:31 pm Post subject:

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Joined: 06 Apr 2008
Posts: 792

Proof of Theorem 4: Let $F,G:[0,1]\times[0,1]\to X$ be homotopies relative to {0,1} between p and q and between r and s respectively. Thus, for all $(t,u)\in[0,1]\times[0,1]$ , we have the following:

F(t,0)=p(t)

Now let $H:[0,1]\times[0,1]\to X$ be defined as follows:

$H(t,u)\ =\ \left\{\begin{array}{ll}F(2t,u)&0\le t\le\frac{1}{2}\\\\G(2t-1,u)&\frac{1}{2}\le t\le1\end{array}$

Note that $F\left(2\left(\frac{1}{2}\right),u\right)=b=G\left(2\left(\frac{1}{2}\right)-1,u\right)$ so H is well defined and continuous at $\left(\frac{1}{2},u\right)$ .

Now let’s check that we have everything else.

$H(t,0)=\left\{\begin{array}{ll}F(2t,0)&0\le t\le\frac{1}{2}\\\\G(2t-1,0)&\frac{1}{2}\le t\le1\end{array}$

$.\hspace{10mm}=\left\{\begin{array}{ll}p(2t)&0\le t\le\frac{1}{2}\\\\r(2t-1)&\frac{1}{2}\le t\le1\end{array}$

$.\hspace{10mm}=p\star r\,(t)$

$H(t,1)=\left\{\begin{array}{ll}F(2t,1)&0\le t\le\frac{1}{2}\\\\G(2t-1,1)&\frac{1}{2}\le t\le1\end{array}$

$.\hspace{10mm}=\left\{\begin{array}{ll}q(2t)&0\le t\le\frac{1}{2}\\\\s(2t-1)&\frac{1}{2}\le t\le1\end{array}$

$.\hspace{10mm}=q\star s\,(t)$

Hence H is a homotopy from $p\star r$ to $q\star s$ . Let’s verify that it’s a homotopy relative to {0,1}.

$H(0,u)=F(2(0),u)=a=p\star r\,(0)=q\star s\,(0)$

$H(1,u)=G(2(1)-1,u)=c=p\star r\,(1)=q\star s\,(1)$

And so it is. The theorem is proved. Cool

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 $\left(\frac{1}{2},u\right)$ . Indeed, it is false that $p\simeq q\ \mbox{and}\ r\simeq s\ \Rightarrow\ p\star r\simeq q\star s$ : the homotopies must be relative to {0,1}. Here is an example to show why.

Let $X=\mathbb{R}^2\setminus\{(0,0)\}$ and let $p:[0,1]\to X$ be as we’ve defined in the opening post of this thread: $p(t)=\left(\cos{\pi t},\sin{\pi t}\right)$ . Now let $q(t)=(\cos{\pi t},2+\sin{\pi t})$ , $r(t)=(-\cos{\pi t},-\sin{\pi t})$ , $s(t)=(-\cos{\pi t},2-\sin{\pi t})$ . Then p and q are homotopic, but not equivalent; similarly for r and s. Now $p\star r$ is the path tracing out the unit circle centred at (0,0) while $q\star s$ traces out the unit circle centred at (0,2). It is clear that $p\star r\not\simeq q\star s$ . The former “encloses” 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” (0,0).
_________________

A problem worthy of attack
Proves its worth by fighting back.
(Piet Hein)

Did You Know?
Fact of the day: Homotopy

Last edited by JaneBennet on Thu Aug 07, 2008 4:12 am; edited 1 time in total

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