Why is it sufficient to show that X has a simply-connected covering space in order to prove that $p_{*}$ is surjective?

Question

Hatcher page 63 states that as $p_{*}$ is always injective, trying to show that it is surjective amounts to finding a simply-connected covering space for X. Can someone possibly explain the logic behind why that is sufficient?

Answer 1

I just happened to have the book handy, but more information would be helpful in the future to those who do not.

Hatcher is not referring to showing that $p_*$ is surjective, he's referring to the function that associates the covering space $(\tilde{X},\tilde{x_0},p)$ (where $p$ is the covering map) to the subgroup $p_*(\pi_1(\tilde{X},\tilde{x_0}))$.

That is, the function is really something like $$\Phi: \mathcal{C}(X,x_0) \to \mathcal{S}(\pi_1(X,x_0)), \Phi(\tilde{X},\tilde{x_0},p) = p_*(\pi_1(\tilde{X},\tilde{x_0})).$$

where $\mathcal{C}(X,x_0)$ is the set of covering spaces of $(X,x_0)$ and $S(\pi_1(X,x_0))$ is the set of subgroups of $\pi_1(X,x_0)$. He wants to answer the question, "is every subgroup of $\pi_1(X,x_0)$ realized this way"? That is, is $\Phi$ surjective? In particular, he wants to know if there's some covering space that the function $\Phi$ sends to the trivial subgroup of $\pi_1(X,x_0)$. We then note that $p_*$ is always injective, so $p_*(\pi_1(\tilde{X},\tilde{x_0}))$ being the trivial subgroup in $\pi_1(X,x_0)$ means that $\pi_1(\tilde{X},\tilde{x_0})$ is the trivial group itself -- meaning that $(\tilde{X},\tilde{x_0},p)$ is simply-connected.

Answer 2

Recall that points of $\tilde{X}$ correspond to homotopy classes of paths based at $x_0$.

Suppose there exists a covering space $\tilde{X}$ that is simply connected. We claim that to every subgroup $G \subset \pi_1(X)$ there exists a cover $X_G$ such that $p_*(\pi_1 (X_G))=G$. First note that $G$ acts on $\tilde{X}$ by the restriction of the monodromy action. We first claim that $\tilde{X}/G$ is a covering space. Note that the resulting space is a quotient by $[\rho] \sim [\alpha]$ iff they have the same endpoint and $[\rho \cdot \overline{\alpha}] \in G$. This is a covering space (as you can check.)

Let $\tilde{x_0}$ "be" the homotopy class of the constant loop at $x_0$. If $[\rho] \in G$, then it lifts to a loop in $X_G$, and likewise if $\alpha \in X_G$ is a loop, then $[p \circ \alpha]$ is a loop in $X$ and $[p \circ \alpha] \in G.$