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?
Consequence of induced homomorphism of covering map being trivial
algebraic-topologygroup-homomorphismhomotopy-theory
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I think the origin of the confusion in the comments is that Munkres denotes loops as maps from intervals, instead of maps from $S^1$. A map $\sigma : [0, 1] \to X$ with $\sigma(0) = \sigma(1) = x_0$ is, however, equivalent to a map $\gamma : (S^1, b_0) \to (X, x_0)$ coming from the universal property of quotient spaces, where the domain circle is the quotient space $[0, 1]/\!\!\sim$, $\sim$ pasting $0$ and $1$ to the point $b_0$. Nevertheless, I'll use the notation you're comfortable with :
$\gamma : [0, 1] \to S^1$ be any loop in $S^1$ based at $\gamma(0) = \gamma(1) = b_0$. As $h_*$ is trivial, $h \circ \gamma : [0, 1] \to X$ based at $h(\gamma(0)) = h(\gamma(1)) = x_0$ can be path-homotoped to the zero loop $e_{x_0}$ at $x_0$. Let $F : [0, 1]\times [0, 1] \to X$ denote the path homotopy between $h \circ \gamma$ and $e_{x_0}$.
The desired nullhomotopy then is $H : S^1 \times [0, 1] \to X$ obtained from the universal property of quotient maps, where $S^1 \times [0, 1]$ is considered as the quotient space $[0, 1] \times [0, 1]/\!\! \sim$ under the identification $(0, x) \sim (1, x)$ for all $x$.
Geometrically, this means we are taking the loop in $(X, x_0)$, and letting the intermediates of the nullhomotopy to be the intermediates of the path-homotopy between $\gamma$ and the zero loop. That's all there is to it, as Qiaochu mentioned in the comments.
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!).
Best Answer
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 $\pi_1(B,b_0)$.