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In the second paragraph of Hatcher's proof, fourth sentence, it says The uniqueness part of (a) implies $\tilde f_0 = \tilde \omega_m$ and $\tilde f_1 = \tilde \omega_n$.

How does the uniqueness part of (a) imply this? If $f_0=\omega_m$ and $f_1=\omega_n$, why is it not automatic that $\tilde f_0 = \tilde \omega_m$ and $\tilde f_1 = \tilde \omega_n$?




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Also, what exactly does that second paragraph show?

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It is 'automatic' by (a). What is not automatic is that $\tilde{\omega}_n(1) = \tilde{\omega}_m(1)$, hence that $n = m$.

To show this, you use the fact that a homotopy lifts to a homotopy. In particular, since $f_t(1)$ is a constant path in $t$, its lifting $\tilde{f}_t(1)$ is still constant in $t$. Then you are done: $$\tilde{\omega}_n(1) = \tilde{f}_0(1) = \tilde{f}_1(1) = \tilde{\omega}_m(1).$$

In the first paragraph he shows that to any loop of $(S^1, x_0)$ we can associate an integer. In the second paragraph he is showing that this number does not depend on the homotopy class of the path. With this he builds the (now well defined) function $$\pi_1(S^1, x_0) \to \mathbb{Z}$$ which is going to be an isomorphism.

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  • $\begingroup$ Why do we need part (a) to say: If $f_0=\omega_m$ and $f_1=\omega_n$ then $\tilde f_0 = \tilde \omega_m$ and $\tilde f_1 = \tilde \omega_n$? $\endgroup$ – Al Jebr Feb 13 at 21:59
  • $\begingroup$ You need a 'theorem' to prove the existence and unicity of liftings of paths. You cannot lift paths along general maps, but (due to (a)) you can do that along covering maps (assuming the spaces are connected). Notice that unicity is up to fixing a point; in the statement (a) it is required to fix the starting point of the path; if you don't fix the starting point then you can get two distinct liftings of the same path. Example: consider the constant path on $S^1$ defined by $t \mapsto 0$, then any constant path on $\mathbb{R}$ defined by $t \mapsto n$ is a lifting of the first path. $\endgroup$ – user574499 Feb 13 at 22:32
  • $\begingroup$ The first path should have been $t \mapsto x_0$. $\endgroup$ – user574499 Feb 13 at 23:05

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