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I need to show the statement in the title, and would like to have some checking on the argument below.

Let $f: S^1 \to X$ be a map, $p \in X$, $e_p: S^1 \to X$ be the constant map, i.e. $e_p(z) = p,\forall z \in S^1$.

The condition $f \simeq e_p$ implies the existence of a map $H: S^1 \times I \to X$ such that $H(z,0) = f(z)$, $H(z,1) = e_p(z) ,\forall z \in S^1$. My intuition tells me that, because of the way $H(z,1)$ is defined, the domain of this map can be re-written as $$ (S^1 \times I)/(S^1 \times \{1\}), $$ which is homeomorphic to $D^2$. So this modified homotopy is the extension we're looking for. But I'm not sure how to make this argument precise and which conditions that I need to check.

ensbana
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1 Answers1

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Consider the map $q$ from $S^1 \times [0, 1]$ to $D^2$ given by

$$ (\theta, t) \mapsto ((1-t) \cos \theta, (1-t) \sin \theta) $$

This is constant on $S^1 \times \{1\}$, and in fact a homeomorphism from $S^1 \times [0, 1] / S^1 \times \{1\}$ to $D^2$, so if we have a function, $$ H : S^1 \times [0, 1] \to X $$ that's constant on $S^1 \times \{1\}$, then $$ H' = H \circ q^{-1} $$ is a continuous map $D^2 \to X$.

Now suppose that $s : S^1 \to X$ is given. By hypothesis, there's a homotopy $$ H: S^1 \times [0, 1] \to X $$ with the property that $H(\theta, 1)$ is constant. Then define $H'$ as above to get a map from $D$ to $X$ that extends $s$.

John Hughes
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  • Thanks. Can you please elaborate how we can verify that $q$ is indeed a homeomorphism from the quotient space to $D^2$? – ensbana Dec 16 '19 at 12:44
  • You see that it's continuous, right? Because it's a composition of continuous maps. You see that it's 1-1 (on the quotient), right? Because $q(a,t) = q(b, s)$ for $t \ne 1$ tells you that $s = t$ and $a = b$, by doing a little algebra. You see it's surjective because...polar coordinates. Now apply this: Theorem: Let $X$ and $Y$ be topological spaces and let $f:X \to Y$. If $X$ is compact, $Y$ is Hausdorff, and $f$ is a continuous bijection, then $f$ is a homeomorphism between $X$ and $Y$. The proof is in most topology texts. – John Hughes Dec 16 '19 at 18:36