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I am working at this problem:

Prove:Every map $S^1 → X$ is homotopic to a constant map$\implies$Every map $S^1 → X$ extends to a map $D^2 → X$.

And I am attempting to understand this solution:

enter image description here

It seems to me that this proof only use the fact that $f_t$ is a homotopy and thus continous. But I cannot see that where do we use the condition that $f$ is homotopic to a constant map. So where does this condition apply? Do we necessarily need it?

Lee Mosher
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user464003
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    There is a typo. That should be $f_t: S^1 \times I \to X$. If $f: X \to Y$ is homotopic to $g: X \to Y$, then there exists a continuous map $H(x,t):=h_t(x)$ where $H: X \times I \to Y$ with.... – Faraad Armwood Jul 31 '17 at 21:12
  • @FaraadArmwood Thanks for pointing it out! – user464003 Jul 31 '17 at 21:14
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    No problem. You can now see where the obvious extension comes from i.e the domain of $H$ which is $S^1 \times I$, a cylinder! – Faraad Armwood Jul 31 '17 at 21:16
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    You can think of the circle at each height $S^1 \times {t}$ as corresponding uniquely to a circle about the origin of radius $rt$. Now if we let $t$ range over $[0,1]$ we get all the circles in the unit disk. I hope this makes sense. – Faraad Armwood Jul 31 '17 at 21:18
  • @FaraadArmwood Yes now I can visualize it, nice interpretation! Thanks a lot! – user464003 Jul 31 '17 at 21:21

1 Answers1

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This fact is used to show that $g(0)=g(0e^{i\theta})=f_0(e^{i\theta})$ is well-defined and does not depend on $\theta$.

mather
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Tsemo Aristide
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