Can anyone give me a proof of why the circle $S^1$ and the closed interval $[0,1]$ are not homotopically equivalent? (Using the basic definition and not the fundamental group!)
Interval and Circle
2
$\begingroup$
algebraic-topology
circles
-
4Why do you need to avoid the fundamental group? This is precisely the sort of question it was invented to answer. – 2012-01-26
-
2To give you an idea: This is equivalent to the fact that $S^1$ is not contractible, which in turn gives a proof of Brouwer's fixed point theorem (since a retract D^2 --> S^1 is the same thing as a nullhomotopy of the identity). Brouwer's fixed point theorem is hard- so this is hard. – 2012-01-26
1 Answers
-2
The interval is simply connected; the circle is not.
-
5$@$Michael: in order for this to qualify as an answer you need to show (i) that simple connectedness is a homotopy invariant and (ii) the circle is not simply connected *without using the fundamental group*. How are you proposing to do this? – 2012-01-26
-
0@PeteL.Clark : I don't know but I'd have thought you couldn't prove the fundamental group is non-trivial without proving that the space is not simply connected. – 2012-01-26
-
0@PeteL.Clark : Is your comment intended simply to mean that I did not prove that the circle is not simply connected? Or did you mean to suggest that such a proof would make use of the fundamental group? If the latter, I'd have some concerns about that. – 2012-01-27
-
1$@$Michael: Yes, it seems that to satisfy the OP one needs a proof of the non simple connectedness which does not use the fundamental group. (It is at least possible to *phrase* the result without the fundamental group: e.g. that the identity map on $S^1$ is not homotopic to a constant map.) – 2012-01-27