# Find a function that makes the following diagram commutes.

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!

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

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 $$\Bbb R^{n+1} -\{0\}$$.
• You need the final modulus of $x$ to be 1, that's why you performed the division. – Nabakumar Bhattacharya Apr 27 '20 at 17:37