Before this is marked as duplicate, I'm gonna say that I understand the proof and what the proposition means mathematically; I just want to know what's wrong with my intuition for it.
(Part of) Hatcher Proposition 1.31 is that the homomorphism of fundamental groups $p_*: \pi_1(\tilde X,\tilde x_0) \rightarrow \pi_1(X,x_0)$ induced by the covering map $p: \tilde X \rightarrow X$ is injective.
Mathematically, all is good, but I'm confused about the picture in my head — my intuition tells me that $\tilde X$ is "bigger" than $X$, and in particular that all loops in $X$ lift to loops in $\tilde X$. Then, given that $\tilde X$ has more loops, shouldn't its fundamental group be bigger? Yet if $p_*$ is injective, it would imply the opposite.
Best Answer
If $p_*$ is injective it means that $\pi_1(\overline{X})$ fits into $\pi_1(X)$ thus it is the latter to be bigger, not the former.
Since $p$ is surjective then $\overline{X}$ is a "bigger" space, but this mean you have more room to contract the loops which could you provide you with non-trivial elements in the fundamental group. Therefore $\pi_1(\overline{X})$ is the smallest between the two groups.
The most intuitive idea I can provide is the following: given the cover $p$ then you could consider $X$ as some sort of quotient of the upper space $\overline{X}$ and $p$ is precisely the quotient map. (You should look for the notion of deck transformation to make this more precise). After taking this quotient, due to the new identifications, you can form new loops which do not lift to loops in $\overline{X}$. Therefore the new group $\pi_1(X)$ is more complicated than the previous $\pi_1(\overline{X})$.
Take as example the exponential map $exp \colon \mathbb{R} \rightarrow S^1$. The first space is contractible, so all loops are homotopically trivial. But the base space is not: because you are identifying all the elements of $\mathbb{R}$ which differ by some integer. Thus the interval $[0,1]$ provides you with a non-trivial element in the quotient space $S^1$ (in this case actually a generator!).