A map $f :S^1\rightarrow X$ is nullhomotopic then $f$ extends to a map $D^2 \rightarrow X$

Question

I know there are other proofs of this question here, but I found a solution and I need some explanation. So we have the homotopy $H$ : $ S^1\times I \rightarrow X$ . This map then induces a map from the the quotient space $(S^1\times I)/(S^1\times {1})$ to $X$. this quotient space is just $D^2$ , so the given map $S^1 \rightarrow X$ extends to a map $D^2 \rightarrow X$. My questions are, how does $H$ induce this map? Which theorem about quotient spaces says that? And then how can we actually say that it extends to the other map? Any help would be much appreciated, thank you

Answer 1

Let $q:X\to \tilde X$ be a quotient map. The universal property of quotient spaces says that whenever there is a map $g:X\to Z$ satisfying $q(a)=q(b)\implies g(a)=g(b)$ then we can uniquely induce a map $h:\tilde X\to Z$ such that $g=h\circ q$.

In your setup, if $q:S^1\times I\to \dfrac{S^1\times I}{S^1\times 1}$ is the quotient map, then $H$ behaves as the map $g$ from the universal property does. This is easy to see this as follows,

Now $q(x,t)=q(x',t')$ either when

1. $(x,t)=(x',t')$ or when

2. $x$ and $x'$ are any points in $S^1$ but $t=t'=1$

In the first case clearly $H(x,t)=H(x',t')$. In the second case, by definition of nulhomotopy you have $H(x,1)=H(x',1)$.

Thus by the universal property of quotient spaces we have a map $H':\dfrac{S^1\times I}{S^1\times 1}\to X$ as desired. In fact we also have $H=H'\circ q$.

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If you are wondering how the map $h$ in the universal property is defined, that is also quite easy to do:

For any $\tilde a\in\tilde X$, consider a preimage $a$ in $X$, that is $q(a)=\tilde a$. Define $h(\tilde a)=g(a)$. This is a well-defined map because if $a'$ is any other preimage then we have $q(a)=q(a')=\tilde a$. But if $q(a)=q(a')$ then $g(a)=g(a')$ so $h$ is well-defined.