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_θ is a path from (1,0) to (−1,0) for each θ ∊ [0,1] and F₀(t) = p(t) and F₁(t) = q(t) for each t ∊ [0,1]. F_θ 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_θ. We say that the path p is homotopic to q in

.

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.

.

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

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.

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

Last edited by JaneFairfax (2008-07-06 06:58:34)

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.

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!

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).

Let

be a path from a to b. The inverse path of p is the path

from b to a where

for all

.