${\Bbb Z}/3{\Bbb Z}$-torsors over $S^1$. I have asked about ${\Bbb Z}/3{\Bbb Z}$-torsors on Visualisation of ${\Bbb Z}/3{\Bbb Z}$-torsors of $S^1$
Therein, I learned that there are two non-trivial ${\Bbb Z}/3{\Bbb Z}$-torsors $T_1$ and $T_2$ over $S^1$, which are both isomorphic to the unique Galois cover $X$ of $S^1$ such that ${\mathrm{Gal}}(X/S^1) \cong {\Bbb Z}/3{\Bbb Z}$. That is, $T_1 \cong T_2 \cong S^1$. However, the projection maps ${\mathrm{pr}}_1 \colon T_1 \to S^1$ and ${\mathrm{pr}}_2 \colon T_2 \to S^1$ to $S^1$ are different from each other.

How can I show that there is no continuous map $f \colon T_1 \to T_2$ such that ${\mathrm{pr}}_1 = {\mathrm{pr}}_2 \circ f$? That is, the base $S^1$ is fixed.

 A: In fact, there is a continuous map $f:T_1\to T_2$ commuting with the projection maps. Let's write $T_1=T_2=S^1=U(1)\subset \mathbb{C}$ with $\operatorname{pr}_1(z) = z^3$ and $\operatorname{pr}_2(z) = z^{-3}$. Then $f(z)=z^{-1}$ is such a map.
So $(T_1,\operatorname{pr}_1)$ and $(T_2,\operatorname{pr}_2)$ are isomorphic in the category of finite covers of $S^1$. But you aren't just thinking about these spaces as covers of $S^1$, you're thinking about them as $\mathbb{Z}/3\mathbb{Z}$-torsors, and we need to take morphisms in the appropriate category to be able to differentiate between them. For $z\in T_1=T_2=U(1)$, the group action here is given for $k\in \mathbb{Z}/3\mathbb{Z}$ by $z\mapsto e^{2k\pi i/3}z$, and we see that $f$ does not respect this action.
At this point you could use some analysis to prove that, in this description of the torsors, our $f$ is actually the only map between $T_1$ and $T_2$ over $S^1$ (modulo composition with the group action), and since it isn't a map of torsors, the two aren't isomorphic as torsors. But the cohomological justification is really what this question is about: if $T_1$ and $T_2$ were isomorphic as $\mathbb{Z}/3\mathbb{Z}$-torsors over$S^1$, their corresponding classes in $H^1(S^1,\mathbb{Z}/3\mathbb{Z})$ would be equal. If you've seen how to construct the Cech cohomology classes corresponding to these torsors, you could do so and show that if there were a map of torsors $f:T_1\to T_2$, the resulting classes will be cohomologous (it will probably take an equal amount of work for you to prove a fairly general statement about when two $G$-torsors $X_1\to Y$ and $X_2\to Y$ represent the same cohomology class, once you understand how to proceed in this case). You should work this out on your own: doing the Cech calculation by hand is the only way to really understand the role of torsors in cohomology. I'll write down a simplified version below that works in this specific context.
Granting your identification $H^1(S^1,\mathbb{Z})\cong \operatorname{Hom}(\pi_1(S^1),\mathbb{Z}/3\mathbb{Z})$, the homomorphism $\phi_T$ corresponding to a $\mathbb{Z}/3\mathbb{Z}$-torsor $\operatorname{pr}:T\to S^1$ is determined by the monodromy action on the fiber of $T$ over your chosen basepoint $p\in S^1$, once we've chosen a trivialization for it (i.e., a bijection $\theta: \mathbb{Z}/3\mathbb{Z}\to \operatorname{pr}^{-1}(p)$ respecting the natural $\mathbb{Z}/3\mathbb{Z}$ action on both sets), as follows: take the the chosen generator $[1]\in \pi_1(S^1)$, lift this loop to a path in $T$ starting from the point $\theta(0)$, and define $\phi_T([1])$ to be the element of $\mathbb{Z}/3\mathbb{Z}$ corresponding to the endpoint of this path under $\theta$. You should now check that if you have a map of torsors $T_1\to T_2$ (again, this respects both the projection maps and the group action), then it carries the fiber over $p$ to the fiber over $p$ and it commutes with the monodromy action on this fiber, and that these two facts imply that the corresponding homomorphisms $\phi_{T_1},\phi_{T_2}\in \operatorname{Hom}(\pi_1(S^1),\mathbb{Z}/3\mathbb{Z})$ are the same.
