0
$\begingroup$

Find a function that makes the following diagram commutes.

Here is the diagram: For $n \geq 0.$ Find a function $r: \mathbb{R}^{n+1} - \{0\} \rightarrow S^n$ filling in the dotted arrow in the following diagram so that the diagram commutes.

$$\require{AMScd} \begin{CD} S^n @>{id_{S^n}}>> S^n\\ @VVV @VVV \\ \mathbb{R}^{n+1} - \{0\} @>{r}>> S^n \end{CD} $$

Where the first vertical arrow is the inclusion map, the second vertical arrow should not exist but I am not skillful in drawing commutative diagrams. and the arrow that has the function $r$ above it should be dotted line because we are searching for this function $f.$

My guess is:

I can take the function to be $r(x) = \frac{x}{|x|}.$ Is my guess correct? I am not sure why I should divide by $|x|,$ or this part should be adjusted to the $n+1$ norm? I do not know.

Any help will be appreciated!

$\endgroup$
2
  • $\begingroup$ What exactly is the commutativity you want? Do you want $r\circ \iota = \operatorname{id}_{S^n}$, where I represent the inclusion by $\iota:S^n\hookrightarrow \mathbb{R}^{n+1}\setminus\{0\}$? $\endgroup$ – MPW Apr 27 '20 at 15:34
  • $\begingroup$ @MPW yes exactly this is what I want. $\endgroup$ – user778657 Apr 27 '20 at 15:39
1
$\begingroup$

I guess by $\vert x\vert$ you mean $\Vert x\Vert$.Then your function is alright. You need to divide by $\Vert x\Vert$ only for sake of well definedness.

This proves $S^n$ is a retract of $\Bbb R^{n+1} -\{0\}$.

It is in fact a strong deformation retract.

$\endgroup$
2
  • $\begingroup$ Could you please show me the well definedness proof? $\endgroup$ – user778657 Apr 27 '20 at 17:28
  • $\begingroup$ You need the final modulus of $x$ to be 1, that's why you performed the division. $\endgroup$ – Nabakumar Bhattacharya Apr 27 '20 at 17:37

Your Answer

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