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
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algebraic-topology
circles
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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
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The interval is simply connected; the circle is not.
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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