23 ottobre 2022

Whitney approximation theorem: $\mathbb{S}^n$ is simply connected for $n \geq 2$

Let me give a proof of the fact that the $n$-sphere $\mathbb{S}^n \subset \mathbb{R}^{n+1}$ is simply connected for $n \geq 2$, which is based on differential topology methods (instead than on the Seifert-Van Kampen theorem). We will use the following two results:

(1) Whitney Approximation Theorem: Every continuous map $f \colon M \to N$ between smooth manifolds is homotopic to a smooth map.
(2) Sard Theorem: Let $f \colon  M \to N$ be a smooth map between smooth manifolds, and let $\operatorname{Crit}(f)$ be the locus of points of $M$ where the differential df has rank $< \dim(N)$. Then the image of $\operatorname{Crit}(f)$ has zero Lebesgue measure in $N$. In particular, if $\dim(M) < \dim(N)$ then $f$ cannot be surjective.
Given a continuous loop $c \colon [0, \, 1] \to \mathbb{S}^n$, we want to show that $c$ is homotopically trivial. By (1), it is not restrictive to assume that $c$ is piecewise smooth. Then, by (2), $c$ cannot be surjective (recall that we are assuming $n \geq 2$).

Now, take a point $p$ not in the image of $c$. Then we can see $c$ as a loop in $\mathbb{S}^n-\{p\}$, which is a contractible space (it is homeomorphic to $\mathbb{R}^n$ via the stereographic projection). So $c$ is homotopically trivial.

Remark. The regularization provided by (1) is essential here. In fact, there exist continuous, non-smooth loops "of Peano type", whose image is the whole sphere.

14 ottobre 2022

Job

Julia Robinson's description of her own job:

Monday: Try to prove theorem
Tuesday: Try to prove theorem
Wednesday: Try to prove theorem
Thursday: Try to prove theorem
Friday: Theorem false

Source: MacTutor

08 ottobre 2022

Principio dei cassetti

Volevo spiegare a mia figlia di otto anni il Principio dei Cassetti, quindi ho cominciato con il solito esempio: "Siccome ci sono più abitanti a Roma che capelli sulla testa di una persona, sicuramente esistono due romani che hanno un numero identico di capelli". E lei: "Beh sì, immagino che in tutta Roma non sia difficile trovare due calvi!"