Why is not $\eta(\tau+1)=\eta(\tau)$ although $\Delta(\tau+1)=\Delta(\tau)$

Question

I am confused with why is not $\eta(\tau+1)=\eta(\tau)$ although $\Delta(\tau+1)=\Delta(\tau)$ and one can define $\eta=\Delta^{1/24}$. Hope this is not a very stupid question. Whereas $\Delta$ is the Discriminant modular form and $\eta$ is the dedekind eta function

Answer 1

The equation $a^n=b^n$ does not imply $a=b$, it only implies $a=\xi b$ for some $n$th root of unity $\xi$.

In particular, $\Delta(\tau+1)=\Delta(\tau)$, or in other words $\eta(\tau+1)^{24}=\eta(\tau)^{24}$, must then imply that $\eta(\tau+1)=\xi\,\eta(\tau)$ where $\xi$ is some $24$th root of unity (note $\xi$ cannot depend on $\tau$ since the $24$th roots of unity are a discrete set). It just so happens that $\xi=\exp(2\pi i/24)$.

The "definition" $\eta=\Delta^{1/24}$ is ambiguous because it doesn't say which $24$th root to extract, and it seems to imply that which $24$th root is being chosen is independent of $\tau$, which is false.