Re: A homotopically trivial path

From: mskirvin@xxxxxxxxx
William Elliot wrote:

Let p:I -> S be a loop at a, ie a path with p(0) = a = p(1).
Let p_a be the constant path at a, ie p(I) = {a}. I = [0,1].

Assume p is homotopic to p_a.
Is p homotopic to p_a with respect to, relative to { 0,1 }?

It's not necessarily the case that your homotopy is rel {0, 1}. For
example, consider loops in the circle S^1. If f and g are any two
loops at 1 (= 1 + 0i, regarding S^1 as the unit circle in the
complexes), then they are homotopic, as can be seen by the homotopy
F:I^2 -> S^1 given by F(s, t) = f((1-t)s)*g(ts). So, F(s, 0) = f(s)
and F(s, 1) = g(s), showing f and g are homotopic.

In particular, any loop in S^1 is homotopic to the constant loop at
1, but since the fundamental group of S^1 is nontrivial, not every
loop is homotopic to the constant loop rel {0, 1}.

Now however if S were contractible, in which case it would contract to
a,
tho not necessarily strongly contract to a, could the conclusion be
made?

| Yes. In fact, if S were just simply connected (a much weaker
| condition than contractible), then the conclusion could be made.

I've seen three equivalent opinions of simply connected

S simply connected when S path connected,
for all a,b in S, p,q path from a to b
==> p homotopic q with respect to { 0,1 }

S simply connected when S path connected,
for all a in S, p,q loop at a
==> p homotopic q with respect to { 0,1 }

S simply connected when S path connected,
for all a in S, p loop at a
==> p homotopic constant loop at a with respect to { 0,1 }

So the question was asking a crutial step in showing contractible spaces
are simiply connected. Yes, contractible spaces are path connected and
any continous function into a contractible space is homotopic to a
constant map.

The trick is to show that loops are homotopic to a constant loop relative
to the endpoints, ie relative to { 0,1 }. I suppect it has something to
do with the domain of a path being compact.

| I'm not sure how much you know about the fundamental group, but
| it is just equivalence classes of loops with f equivalent to g
| if f is homtopic to g rel {0, 1}. So, if the fundamental group
| is trivial (this is what simply connected means), every loop is
| homotopic to every other loop rel {0, 1}. If you haven't seen it
| before, it is not too difficult to prove that the fundamental group
| of a contractible space is trivial.

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Relevant Pages

  • Re: A homotopically trivial path
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  • Re: Simply Connected
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  • Re: Simply connected, Not contractible
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  • Re: A homotopically trivial path
    ... It's not necessarily the case that your homotopy is rel ... In particular, any loop in S^1 is homotopic to the constant loop at 1, ... that any two loops centered at the identity in a topological monoid are ... out there with nontrivial fundamental group that could furnish further ...
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  • Re: Simply Connected
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