From Hatcher: Classification of Covering Spaces In the last paragraph, is the function he refers to assigning "different" covering spaces to the subgroup of $\pi_1(X,x_0)$? Also, I did not understand the last sentence. How does knowing that $p_*$ is always injective, amount to asking whether X has a simply-connected covering space? I understand that for X's covering space to be simply-connected it must have a trivial fundamental group and be path-connected.


With respect to your first question, the function he refers to is assigning subgroups of $\pi_1(X,x_0)$ to different covering spaces $(\tilde X,\tilde x_0)$ of $(X,x_0)$.

Since $p_*$ is always injective, by the first isomorphism theorem, $p_*(\tilde X,\tilde x_0)$ is isomorphic to a subgroup of $\pi_1(X,x_0)$. Suppose you ask the question "Does every subgroup of $\pi_1(X,x_0)$ have a corresponding covering space?" Well, the answer to your question depends on whether there is a covering space of $(X,x_0)$ that corresponds to the trivial subgroup (the $0$ subgroup). Thus, we first need to answer "Does $(X,x_0)$ have a simply-connected covering space?" as such a covering space corresponds to the $0$ subgroup of $\pi_1(X,x_0)$.

  • $\begingroup$ @ AOrtiz I did not understand the explanation to my second question. Can you please expand a little more on the "connection" between the two questions? Is the 0 subgroup the trivial subgroup? $\endgroup$ – cbro Jul 20 '17 at 2:37
  • $\begingroup$ @cbro I have updated my answer $\endgroup$ – Alex Ortiz Jul 20 '17 at 2:40

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