# If $h:S^1\to X$ is homotopic to a point, then h can be extended to $H:D^2\to X$.

Let $h:S^1\to X$ be a continuous map. Show if $h$ is homotopic to a point, then $h$ can be extended to a continuous map $H:D^2\to X$.

Intuitively is a little bit obvious, let $F:S^1\times I\to X$ a homotopy, where $F(x,0)=h(x)$ and $F(x,1)=x_0 \in X$ because you can pass $h$ to the quotient, if $q:S^1\times I\to S^1\times I/S^1\times \{1\}$ is the quotient map so there is a continuous map $\bar h:S^1\times I/S^1\times \{1\}\to X$, such that $h=\bar h\circ q$.

There are also homeomorphisms between $D^2$ and $S^1\times I/S^1\times \{1\}$ and $S^1\times \{0\}$ and $S^1$. We know also that $F$ restricted to $S^1\times \{0\}$ is $f(x)$. Even with these informations I can't prove the question, I think we have to compose these maps to have a map $H$, but I can't prove that the restriction of $H$ in $S^1$ is $f$. The question is more subtle than I thought at the first sight.

I need help

Thanks

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I think you are alread there and didn't notice. The $S^1\subseteq D^2$ is just the $S^1\times \{0\}\subseteq S^1\times I$. –  Hagen von Eitzen Nov 11 '12 at 22:21
Why not just let $H(tp)=F(p,1-t)$ for $p\in S_1$ and $t\in[0,1]$? –  Brian M. Scott Nov 11 '12 at 22:48
@BrianM.Scott Is that correct? I don't believe, very simple and elegant solution, thank you very much. –  user42912 Nov 12 '12 at 1:47
@user42912. I'd like to remark that the correct formulation would be "$h$ is homotopic to a constant map", or briefly "$h$ is nullhomotopic", although it is clear what you mean. –  Stefan Hamcke Nov 12 '12 at 14:14
You’re welcome. –  Brian M. Scott Nov 12 '12 at 16:36

Suppose that any map $f: S^1 \rightarrow X$ is homotopic to the constant map $c$ that sends every point of $S^1$ to some $x_0 \in X$. Suppose the homotopy is given by a map $F: S^1 \times I \to X$ such that $F(x,0) = x_0$, $F(x,1) = f(x)$ for all $x \in S^1$. Define $g : D^2 \to X$ by
$$g(x) = \begin{cases} x_0,&\text{if} \hspace{2mm} 0 \leq ||x|| \leq \frac{1}{2} \\ F(x/||x||,2 - 2||x||),& \text{if} \hspace{2mm} \frac{1}{2} \leq ||x|| \leq 1 \end{cases}.$$
The $x/||x||$ appearing in $F$ guarantees that every point in $D^2$ is sent to the unit circle $S^1$ so it makes sense to apply $F$ to such a point. The continuity of $g$ is evident from the fact that $F(x/||x||,2-2\frac{1}{2}) = F(x,1) = x_0$. Clearly $g$ extends $f$ because if $||x||=1$, $g(x) = F(x,0) = f(x)$.
you made a mistake $F(x,1)=f(x)$ –  user42912 Nov 15 '12 at 2:10