[Math] Every map $S^1 \rightarrow X$, extends to a map $D^2 \rightarrow X$

algebraic-topologygeneral-topology

I am reviewing some solutions for past papers for an upcoming exam, and I am a bit confused with the solution for this question. I am hoping someone can shed some light for me:

I want to prove that:

Every map $S^1 \rightarrow X$ is homotopic to a constant map
$\implies$ Every map $S^1 \rightarrow X$, extends to a map $D^2 \rightarrow X$

Now the solutions state that:

if $h: S^1 \times I \rightarrow X$ is a homotopy taking a map $f:S^1
\rightarrow X$ to a constant map, then $h$ factors through the
quotient $S^1 \times I/S^1 \times \{1\}$

i.e. $h$ is equal to the composition $S^1 \times I \rightarrow S^1
\times I/S^1 \times \{1\} \rightarrow X$

where the first map is the quotient map.

The pair $(S^1 \times I/S^1 \times \{1\}, S^1 \times I)$ is
homeomorphic to $(D^2,S^1)$, and the map above gives a map $D^2
\rightarrow X$ such that the restriction to $S^1$ is equal to $f$.

I'm not quite sure I understand the quotient $I/S^1$. I know $S^1$ can be constructed as a quotient of the interval $[0,1]$, but what is $I/S^1$? Do we just 'quotient' off all the points in the interval $[0,1]$ except for the endpoints?

I'm also not understanding how the map: $S^1 \times I \rightarrow S^1
\times I/S^1 \times \{1\} \rightarrow X$

gives a map from $D^2 \rightarrow X$ where the restriction to $S^1$ is equal to $f$.

Would greatly appreciate any help to understand this. Thanks so much.

Best Answer

As nactusraid said in the comments, the parentheses are incorrect. The idea is that if you collapse one boundary component of an annulus ($S^1 \times I$) to a point, you get a space that is homeomorphic to $D^2$ and the homeomorphism takes the other boundary component to $\partial D^2 = S^1$.

There is an easy point set topology theorem that says that if you have a continuous function $f \colon X \to Y$ and a quotient map $g \colon X \to Z$ such that $f$ is constant on the fibers of $g$, there exists a map $h \colon Z \to Y$ with $f = h \circ g$.

Applying this to your situation, you have a homotopy $h \colon S^1 \times I \to X$ between $f$ and a constant map. The above theorem gives you a continuous map from $g \colon (S^1 \times I) / (S^1 \times \{1\}) \to X$. Since $h$ restricted to $S^1 \times \{0\}$ is equal to $f$, you see that $g$ restricts to $f$.

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