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Stream: learning: questions

Topic: 2-sphere isn't an open subset of a cartesian space

Jean-Baptiste Vienney (Jun 30 2024 at 00:13):

In this nlab entry: Karoubi envelope, they say that the category of smooth manifold is (equivalent, isomorphic? to) the idempotent-splitting completion of the category of open subspaces of cartesian spaces. In remark 1.2, it is said that the $2$-sphere can be represented as say $(\mathbb{R}^3\backslash \{0\},e)$ where $e$ is the endomap of $\mathbb{R}^3\backslash \{0\}$ defined by $e(x)=\frac{x}{||x||}$.

It makes a lot of sense to me since in the usual representation of the category of smooth manifolds, we have this splitting of $e$:
$\mathbb{R}^3\backslash \{0\} \overset{r}{\rightarrow} S^2 \overset{s}{\rightarrow} \mathbb{R}^3\backslash \{0\}$
where $r(x)=\frac{x}{||x||}$, and $S^2$ is seen as the sphere of radius one centered at $0$ in $\mathbb{R}^3$, and $s$ is the inclusion.

It is said in this same remark that the 2-sphere isn't (diffeomorphic) to an open subset of a cartesian space.

I don't know how to prove this.

Questions:

• How to prove that the $2$-sphere isn't diffeomorphic to an open subset of a cartesian space?
• Is the stronger statement that the $2$-sphere isn't homeomorphic to an open subset of a cartesian space true?
• What about the other $n$-spheres?

Eric M Downes (Jun 30 2024 at 00:37):

I don't remember much about diffeomorphisms, sadly.

By "cartesian space" you mean $\mathbb{R}^n$ right? If so I can help for parts 2 and 3. Can you prove that $n$-spheres are compact?

(Hint: can you find a continuous injective embedding $f:S^n\hookrightarrow\mathbb{R}^{n+1}$ such that $f(S^n)$ is closed and bounded?)

(If you mean something else by cartesian space please link to that.)

Jean-Baptiste Vienney (Jun 30 2024 at 00:41):

Eric M Downes said:

By "cartesian space" you mean $\mathbb{R}^n$ right?

Yes, exactly.

Jean-Baptiste Vienney (Jun 30 2024 at 00:42):

And diffeomorphic $\Rightarrow$ homeomorphic, so not homeomorphic $\Rightarrow$ not diffeomorphic.

Jean-Baptiste Vienney (Jun 30 2024 at 00:44):

Eric M Downes said:

(Hint: can you find a continuous injective embedding $f:S^n\hookrightarrow\mathbb{R}^{n+1}$ such that $f(S^n)$ is closed and bounded?)

Yes, the usual representation of $S^n$ is as the sphere of radius $1$ centered at $0$ in $\mathbb{R}^{n+1}$. With this representation, the embedding is just an inclusion.

Eric M Downes (Jun 30 2024 at 00:47):

Nice. So is it compact then?

Eric M Downes (Jun 30 2024 at 00:48):

(and is compactness a topological invariant?)

Jean-Baptiste Vienney (Jun 30 2024 at 00:48):

Ok, $S^n$ is closed and bounded in $\mathbb{R}^{n+1}$, thus it is compact. But is it true that an open subset of cartesian space is never compact?

Eric M Downes (Jun 30 2024 at 00:50):

Well, compact = closed & bounded by Heine-Borel right? And under the usual topology of $\mathbb{R}^n$ there are very few clopen sets...

Jean-Baptiste Vienney (Jun 30 2024 at 00:51):

Hmm, an open subset of a cartesian space is homeomorphic to the open ball of radius $1$ centered at $0$ (through $x \mapsto \frac{x}{1-||x||}$ whose inverse is $x \mapsto \frac{x}{1+||x||}$)which is not compact because you can define a sequence which converges to the border.

Eric M Downes (Jun 30 2024 at 00:53):

Be careful though..... is $(0,1)\cup(2,3)$ open?

Jean-Baptiste Vienney (Jun 30 2024 at 00:53):

Yes?

Jean-Baptiste Vienney (Jun 30 2024 at 00:54):

Why do I need to be careful?

Eric M Downes (Jun 30 2024 at 00:55):

Is simple-connectedness a topological invariant?

Eric M Downes (Jun 30 2024 at 00:55):

eg preserved by homeomorphisms

Jean-Baptiste Vienney (Jun 30 2024 at 00:56):

Jean-Baptiste Vienney said:

Hmm, an open subset of a cartesian space is homeomorphic to the open ball of radius $1$ centered at $0$ (through $x \mapsto \frac{x}{1-||x||}$ whose inverse is $x \mapsto \frac{x}{1+||x||}$)which is not compact because you can define a sequence which converges to the border.

Sorry, I don't think the first fact is true. I was thinking "$\mathbb{R}^n$ is homeomorphic to the open ball of radius $1$ centered at $0$".

Eric M Downes (Jun 30 2024 at 00:56):

yes! thats what I was getting at. Nice work.

Jean-Baptiste Vienney (Jun 30 2024 at 00:57):

Yeah, it is false, since an open set isn't necessarily connected.

Jean-Baptiste Vienney (Jun 30 2024 at 00:57):

Eric M Downes said:

Well, compact = closed & bounded by Heine-Borel right? And under the usual topology of $\mathbb{R}^n$ there are very few clopen sets...

Ok, this is the way to go I guess.

Jean-Baptiste Vienney (Jun 30 2024 at 01:03):

Ok, if $U$ is a clopen set in $\mathbb{R}^n$ which is neither $\emptyset$ nor $\mathbb{R}^n$, then $\mathbb{R}^n \backslash U$ is also clopen. So $\mathbb{R}^n$ is a disjoint union of nonempty open sets, thus $\mathbb{R}^n$ is not connected. But $\mathbb{R}^n$ is connected since it is path-connected.

Jean-Baptiste Vienney (Jun 30 2024 at 01:10):

To sum up:

• The $n$-sphere is compact as a bounded closed subset of $\mathbb{R}^{n+1}$.
• If the $n$-sphere is homeomorphic to an open subset $U$ of an $\mathbb{R}^p$, then $U$ is compact, thus closed, thus a clopen. So $U$ is either $\mathbb{R}^p$ or the empty set. The cardinality is preserved by homeomorphism, thus $U$ is necessarily $\mathbb{R}^p$.

Now, we have reduced the problem to the one of proving that the $n$-sphere isn't homeomorphic to an $\mathbb{R}^p$.

Jean-Baptiste Vienney (Jun 30 2024 at 01:11):

Ok, I think I know how to finish.

Jean-Baptiste Vienney (Jun 30 2024 at 01:19):

I think you must show that the $\mathbb{R}^p$ is contractible but the $n$-sphere is not.

Eric M Downes (Jun 30 2024 at 01:29):

I would say they are both contractible to different things, but yes that is sufficient.

I mean, you don't need homotopy though. If you know what homeomorphisms preserve, you can see it directly, which is maybe more categorical? Anyway, I think you're good; I gotta go though, so good luck!

Eric M Downes (Jun 30 2024 at 01:32):

Oh yeah sorry forgot that people use "contractible" to mean "contractible to a point" instead of say "contractible to a loop" etc. Sorry, I always found that confusing and must have chosen to forget it lol.

Jean-Baptiste Vienney (Jun 30 2024 at 01:35):

Anyway, I wanted to say that the $n$-sphere is not contractible (if $n \ge 1$) because $H^n(S^n) \cong \mathbb{Z}$.

Jean-Baptiste Vienney (Jun 30 2024 at 01:38):

Oh, maybe it's overkill.

Jean-Baptiste Vienney (Jun 30 2024 at 01:39):

It's easier to show that $\mathbb{R}^n$ is not compact I think.

Jean-Baptiste Vienney (Jun 30 2024 at 01:40):

It's not compact because it is not a bounded subset of itself.

Jean-Baptiste Vienney (Jun 30 2024 at 01:42):

Ok, so you just have to say this:

Let $n \ge 1$.

• The $n$-sphere is compact as a bounded closed subset of $\mathbb{R}^{n+1}$.
• If the $n$-sphere is homeomorphic to an open subset $U$ of an $\mathbb{R}^p$, then $U$ is compact, thus closed, thus a clopen. So $U$ is either $\mathbb{R}^p$ or the empty set. The cardinality is preserved by homeomorphism, thus $U$ is necessarily $\mathbb{R}^p$. Thus $\mathbb{R}^p$ is compact. But $\mathbb{R}^p$ is not compact if $p \ge 1$ because it is a subset of $\mathbb{R}^p$ which is not bounded. Hence, $p=0$. Thus $S^n$ is homeomorphic to a point which is false because it is an infinite set.

Jean-Baptiste Vienney (Jun 30 2024 at 01:43):

It was not that hard ahaha.

Todd Trimble (Jun 30 2024 at 05:54):

In this nlab entry: Karoubi envelope, they say that the category of smooth manifold is (equivalent, isomorphic? to) the idempotent-splitting completion of the category of open subspaces of cartesian spaces.

The idempotent-splitting completion is determined up to equivalence, certainly.