0
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ The map on the right is assumed to be the identity as well? $\endgroup$ – PrudiiArca Apr 27 '20 at 18:10
  • $\begingroup$ @PrudiiArca sure $\endgroup$ – user778657 Apr 28 '20 at 3:09
1
$\begingroup$

$\Bbb S^0 =\{\pm 1\}$ 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.

$\endgroup$
8
  • $\begingroup$ I do not understand your first line ... why we are sure that "there is no surjective continuous map $\mathbb{D}^1 \rightarrow \mathbb{S}^0$" could you explain this please? $\endgroup$ – user778657 Apr 28 '20 at 3:19
  • 1
    $\begingroup$ The image of a connected space under a continuous map is again connected. Thus every map $\Bbb D^1 \rightarrow \Bbb S^0$ is constant and in particular does not hit either 1 or $-1$. $\endgroup$ – PrudiiArca Apr 28 '20 at 6:08
  • $\begingroup$ I agree with your first statement but I do not understand your second statement .... why every map is constant ? why it does not hit either 1 or -1? $\endgroup$ – user778657 Apr 28 '20 at 6:21
  • 1
    $\begingroup$ Because $\{-1,+1\}$ has two connected components, namely $\{-1\}$ and $\{1\}$. Any continuous map from a connected space into a disconnected space can only hit one of the connected components $\endgroup$ – PrudiiArca Apr 28 '20 at 6:37
  • 1
    $\begingroup$ Oh right. Sorry for creating this misunderstanding. In German this might have been clearer (or similarly unclear I guess), maybe I mistranslated :( $\endgroup$ – PrudiiArca Apr 28 '20 at 19:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy