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Proving a continuous function in S1 is homotopic to a constant function in X

MikeD

New member
Feb 16, 2012
2
Hi,

Ive been going over some Topology past papers for a course I'm currently doing and this question really has me stumped. Any help you can offer I'd really appreciate...

Let $\sim$ be the equivalence relation on $S^1$ x $I$ which identifies all the points in $S^1$ x $\left \{ 1 \right \}$, and let

$W = (S^1$ x $I ) / \sim$

be the identification space. Construct a homeomorphism $\phi : W \to D^2$ such that $\phi[z, 0] = z \in S^1$ for all $z \in S^1$. Write down a formula for the inverse homeomorphism $\phi^{-1}: D^2 \to W$.

Then use this solution to prove that a continuous function $f : S^1 \to X$ is homotopic to a constant function iff $f$ has an extention to a continuous function $F : D^2 \to X$. This should include explicit formulae relating $F$ and a homotopy to $f$ to a constant map.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
Hi,

Ive been going over some Topology past papers for a course I'm currently doing and this question really has me stumped. Any help you can offer I'd really appreciate...

Let $\sim$ be the equivalence relation on $S^1$ x $I$ which identifies all the points in $S^1$ x $\left \{ 1 \right \}$, and let

$W = (S^1$ x $I ) / \sim$

be the identification space. Construct a homeomorphism $\phi : W \to D^2$ such that $\phi[z, 0] = z \in S^1$ for all $z \in S^1$. Write down a formula for the inverse homeomorphism $\phi^{-1}: D^2 \to W$.

Then use this solution to prove that a continuous function $f : S^1 \to X$ is homotopic to a constant function iff $f$ has an extention to a continuous function $F : D^2 \to X$. This should include explicit formulae relating $F$ and a homotopy to $f$ to a constant map.
The geometric idea here is to think of $W$ as a cone of height 1 with base $S^1$, and to think of $D^2$ as the disc whose perimeter is the base of the cone. Then $\phi$ should be the map that projects a point of the cone vertically down to the point of the disc that is directly beneath it. Translate that idea into an algebraic formula for $\phi$.