In Hatcher's algebraic topology, page 63, it reads:

"The Galois correspondence arises from the function that assigns to each covering space $\ p: (\bar{X}, \bar{x_{0}}) \rightarrow (X, x_{0}) $ the subgroup $\ p_{*}(\pi_{1}(\bar{X}, \bar{x_{0}})) $ of $\ \pi_{1}(X, x_{0})$. First we consider whether this function is surjective. That is, we ask whether every subgroup of $\ \pi_{1}(X, x_{0})$ is realised as $\ p_{*}(\pi_{1}(\bar{X}, \bar{x_{0}})) $ for some covering space $\ p: (\bar{X}, \bar{x_{0}}) \rightarrow (X, x_{0}) $ . In particular we can ask whether the trivial subgroup is realized. Since $\ p_{*}$ is always injective, this amounts to asking whether $\ X $ has a simply-connected covering space. Answering this will take some work. "

Here $\ p_{*} $ is the homomorphism induced by the continuous function p, Hopefully the rest is clear.

My question concerns the last part. If I have a simply connected covering space thens its fundamental group is trivial and thus any homomorphism acting on this group must have an image that is the trivial subgroup. This is true regardless of whether the homomorphism is injective or not, it would always be true. So why does Hatcher say "Since $\ p_{*}$ is always injective" when that shouldnt matter?

I do appreciate that the injectivity of the homomorphism makes the condition necessary but it is sufficient without it.

  • 2
    $\begingroup$ It does matter, since if $p_*$ wasn't injective, you may get $p_*(G)=1$ for $G$ non-trivial, making you falsely think there is a cover with trivial fundamental group. $\endgroup$ – Pedro Tamaroff Apr 12 '18 at 13:12
  • $\begingroup$ I appreciate that, as i said in the last paragraph the injectivity makes having a simply connected covering space a neccesary condition for the trivial group to be realised, but Iam saying it is sufficient to have the existence of a simply connected covering space to imply that the trivial subgroup is realised without the injectivity restriction. $\endgroup$ – Hugh Kinnear Apr 12 '18 at 13:32
  • $\begingroup$ That's a fair observation, but you're saying that "there is a cover with simply connected fundamental group" implies "there is a trivial subgroup in the image of $p_*$". Hatcher wants to remark that this is an iff. $\endgroup$ – Pedro Tamaroff Apr 12 '18 at 13:37
  • $\begingroup$ Ah okay, I personally don't think it is made clear that it is an iff in the text but thank very much for clearing that up, it is appreciated. $\endgroup$ – Hugh Kinnear Apr 12 '18 at 13:38
  • $\begingroup$ What gives it away here is "amounts to", which you can safely replace by "is equivalent to". $\endgroup$ – Pedro Tamaroff Apr 12 '18 at 13:52

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