Consequence of induced homomorphism of covering map being trivial?

Question

Suppose $p: A \rightarrow B$ is a covering map and it has been shown that the induced homomorphism $p_*$ between the two fundamental groups of $A$ and $B$ must be trivial. Can someone provide some intuition as to why the fundamental group of $B$ must be trivial?

Answer 1

This is clearly not true. The map $p:\mathbb{R}\to S^1$ defined by $p(t)=e^{2\pi it}$ is a covering map. Since $\mathbb{R}$ is simply connected we have $\pi_1(\mathbb{R}, 0)\cong\{e\}$, and so $p_*$ is a trivial homomorphism. However, $\pi_1(S^1,1)\cong\mathbb{Z}$, it is not trivial.

What is true is that if $p:E\to B$ is a covering map with $p(e_0)=b_0$ then $p_*:\pi_1(E,e_0)\to\pi_1(B,b_0)$ is injective. Indeed, if $[f]_E\in Ker(p_*)$ then $p\circ f$ is homotopic to the constant loop $b_0$ in $B$. Note that $f$ is exactly the (unique) lifting of $p\circ f$ to $E$ which begins at the point $e_0$. Liftings of homotopic paths are homotopic. So if $[p\circ f]_B=[e_{b_0}]_B$ then $[f]_E=[e_{e_0}]_E$. So the kernel of $p_*$ is trivial.

So in particular, if $p_*$ is trivial then $\pi_1(E,e_0)$ is trivial. Anyway, the fundamental group of the covering space is "smaller", it is isomorphic to a subgroup of $\pi_1(B,b_0)$.