# Showing that the dotted arrow can not be filled in with a continuous function making the diagram commutes.

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?

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

$$\Bbb S^0 =\{\pm 1\}$$ 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.
• 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? – user778657 Apr 28 '20 at 3:19
• 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 $-1$. – PrudiiArca Apr 28 '20 at 6:08
• 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 – PrudiiArca Apr 28 '20 at 6:37