you need to prove that $H^1(S^1)$ is torsion-free (it's clear that it's finitely generated). Note that $\pi_n(S^1) = 0$ for $n\geq 2$ by the covering $\Bbb R \to S^1$ and LES of homotopy groups for fibrations, so that $S^1 = K(\pi_1(S^1),1)=K(H_1(S^1),1))$ (by Hurewicz + $\pi_1$ of top. groups is abelian). If there was any torsion in $\pi_1(S^1)=H_1(S^1)$, then the cohomological dimension of $\pi_1(S^1)$ would be infinite,
contradicting the fact that it is finite since $S^1$ is a finite-dimensional manifold

"We can say the same about Tymoczko’s worries concerning the sophistication
of some mathematical approaches used by Mazzola to delve into the intricacies
of musical phenomena. From his examination of Mazzola’s The Topos of Music,
he concludes [11] that Mazzola is tacitly asserting “If you cannot learn algebraic
geometry [...], then you have no business trying to understand Mozart”. Such
a phrase is similar to “If you cannot learn probability and statistics, then you
have no business trying to count the population of a city”. Of course you