This Theorem is assumed to be taken for granted:
$S^1$ is not contractible.
Now I want to use it to Show that the dotted arrow can not be filled in with a continuous function making the following diagram commutes.
$$\require{AMScd} \begin{CD} S^0 @>{id_{S^0}}>> S^0\\ @VVV @VVV \\ D^1 @>{?}>> S^0 \end{CD}$$
where the arrow below $?$ should be a dotted arrow because we are searching for this function. And I am not skillful in drawing commutative diagrams this is why I draw $S^0$ 2 times because I do not know how to draw one dotted arrow coming out of $D^1$ going directly to $S^0$ my bad. Then my job is to show that there can be no such function $?$
My thoughts:
I know that if there were such function, call it $r.$ And if I call the function from $S^0$ to $D^1$ say $j$ then by the commutativity of the diagram I would have $r \circ j = id_{S^0}.$ But how this would contradict that $S^1$ being not contractible ? I know that contractible means homotopically equivalent to the one point space $*.$
Any help with directing my thoughts in the right direction please?
$\Bbb S^0 =\{\pm 1\}$ is disconnected and $\Bbb D^1 =[0,1]$ is connected, hence there is no surjective continuous map $\Bbb D^1 \rightarrow \Bbb S^0$ and in particular no retraction to $\Bbb S^0$.
More interesting is the question, whether $\Bbb S^1$ is a deformation retract of $\Bbb D^k$ for any $k\geq 1$, ie. if there is an embedding $m:\Bbb S^1 \rightarrow \Bbb D^k$ and a retraction $r:\Bbb D^k \rightarrow \Bbb S^1$ satisfying $rm = \operatorname{id}_{\Bbb S^1}$ and $mr \sim \operatorname{id}_{\Bbb D^k}$. However, as deformation retracts constitute homotopy equivalences and all $\Bbb D^k$ are contractible, ie. homotopy equivalent to $*$, the knowledge that $\Bbb S^1$ is not contractible, answers this question to the negative as well.