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#1 2008-07-03 13:27:46

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Paths

Let

be defined by
. Then p is a path in
from (1,0) to (−1,0). In general: 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] → X such that p(0) = a and p(1) = b. In our example, our path is the upper semicircle of the unit circle centred at the origin. smile

Last edited by JaneFairfax (2008-07-03 13:39:02)


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#2 2008-07-03 22:35:25

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Paths

Let

where q(t) = −p(t) for each t ∈ [0,1] be another path in
(the lower semicircle of the same unit circle). Suppose for each θ ∈ [0,1], we have a function
defined by
.

Note that F[sub]θ[/sub] is a path from (1,0) to (−1,0) for each θ ∊ [0,1] and F[sub]0[/sub](t) = p(t) and F[sub]1[/sub](t) = q(t) for each t ∊ [0,1]. F[sub]θ[/sub] is a semiellipse with semimajor axis 1 and semiminor axis |1−2θ| (except when θ = ½, when it’s a straight-line path).

Now, what happens when θ varies continuously from 0 to 1? Yes, the path p transforms continuously to q via the various intermediate paths F[sub]θ[/sub]. We say that the path p is homotopic to q in

. smile


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#3 2008-07-04 08:53:18

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Paths

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

to mean that F is a homotopy from f to g. smile


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#4 2008-07-05 00:56:52

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Paths



And we write

. F can be seen as “fixing” the subset A while deforming f continuously into g. In particular, if p and q are paths, we can have a homotopy H which “fixes” the start and end points – i.e.
. smile


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#5 2008-07-05 08:04:17

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: Paths

I see you're starting to get into algebraic topology.  Is this self study?  What book are you using?


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#6 2008-07-05 08:51:29

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: Paths

Yes, it’s mainly self-study – but any help from anybody will be greatly appreciated. tongue

I’m using A First Course in Algebraic Topology (2005) by B.K. Lahiri (one of the books I bought at Waterstone’s next to University College London on my visit to Central London the other day). I hate to say that it’s not very well written (though it’s a second edition and the author says mistakes in the original edition have been corrected). Hopefully they are not serious mistakes, and a person like me should have little or no problem seeing them for what they are. tongue

I’m also supplementing my reading with whatever material I can find online.

Last edited by JaneFairfax (2008-07-05 08:58:34)


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#7 2008-07-05 12:23:24

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Paths

By the way, MathsIsFun, if you need a page on algebraic topology on the website, I can write one up for you – up to and including the defintion of fundamental group. big_smile


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#8 2008-07-13 02:06:05

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Paths

Let a, b, c be points in a topological space X. Suppose

are paths from a to b and
are paths from b to c. First we define the product path of two paths,
:

Basically,

is a path from a to c via b. smile

Then we have this reault:

It must be noted that it is important for the homotopies to be relative to {0,1}. If p and q (likewise r and s) are only homotopic, not homotopic relative to {0,1}, then the product paths

and
need not be homotopic! shame

Here is an example to show why. Let

and define paths as follows:




Then p and q (likewise r and s) are only homotopic, not homotopic relative to {0,1} – and the product paths

and
are definitely not homotopic. The former is the unit circle centred at (0,0) and the latter the unit circle centred at (0,2) – as (0,0) is not in X, there is no way a loop enclosing (0,0) can be continuously deformed in the Cartesian plane into a loop that does not enclose (0.0). neutral


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#9 2008-07-13 09:47:22

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Paths

Let

. The null path at a is the constant path at a – i.e. the path
where
for all
. smile


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#10 2008-07-13 12:54:20

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: Paths

Let

be a path from a to b. The inverse path of p is the path
from b to a where
for all
. smile


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#11 2008-07-23 21:04:31

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Paths



Last edited by JaneFairfax (2008-07-23 21:14:02)


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#12 2008-07-26 04:41:27

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Paths


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#13 2008-07-26 09:42:28

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Paths



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A: Click here for answer.

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#14 2008-08-07 07:08:23

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Paths

Yearning.gif

big_smile


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