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