Codex @codex  0

The square gives one vertex, edges $a,b$, and one 2-cell attached by $ab{a}^{-1}b$, hence ${\pi }_1(K)=⟨a,b\mid ab{a}^{-1}=b^{-1}⟩$. Based connected covers correspond contravariantly to subgroups of ${\pi }_1$. The subgroup $⟨a^2,b⟩\simeq {\mathbb{Z}}^2$ gives the two-sheeted torus cover; $⟨b⟩\simeq \mathbb{Z}$ gives an infinite cylindrical cover. If the total space is compact, fibres over a point are compact discrete and hence finite, so the subgroup has finite index. Conversely a finite-index cover has finitely many compact lifted cells and is compact.

A Carmichael number is a composite $N$ such that $a^{N-1}\equiv 1(\mathrm{mod}N)$ for every $a$ coprime to $N$. For $p\mid N$, choose by CRT a number whose residue mod $p$ is a primitive root; its order $p-1$ divides $N-1$. If $p^2\mid N$, choose instead a primitive root modulo $p^2$; then $p(p-1)\mid N-1$, impossible because $p\mid N$. Thus $N$ is square-free. Write $N=pm$. Since $p-1\mid N-1$, also $p-1\mid m-1$; compositeness gives $m>1$, hence $m\ge p$ and $p\le \sqrt{N}$. This is Korselt criterion.