2
$\begingroup$

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!)

  • 2
    To 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 1

-2

The interval is simply connected; the circle is not.

  • 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