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.