Liftings in Topology

Question

I am wondering about this: Assume you have a class $[f] \in \pi_1(B,b_0)$ and a covering map $p:E \rightarrow B$. Now, I know that if you take any two paths $g,h \in f$ that are homotopic and they have liftings $\hat{g} , \hat{h}$, then if $E$ is not simply connected it might happen that the liftings are not closed, correct? But since we know that homotopy is inherited, we have that the liftings are homotopic, too. So doesn't this mean that if it fails for any $f \in [f]$ that the lift is closed, then it also fails for all other elements of the class?-Due to the fact that homotopy of paths requires them to have the same endpoints. So nobody can have a closed lift in this class?

Now, if this is true: Can I conclude from this that if I have a class with an element $f$ in it that has a lift that is not closed, then $[f]$ cannot be the unit element in the group $\pi_1(B,b_0)$?- This is merely a direct consequence of the first question ( in case that the answer is yes)

Answer 1

Yes, everything you said is correct. Here's one way to see how the first point is correct (then the second question is, as you say, a direct consequence):

The fiber $p^{-1}(b_0)$ is discrete. Fix a lift $\hat{b}_0 \in p^{-1}(b_0)$, I assume all the lifts start at this point. If $g,h \in f$, fix a homotopy $H : [0,1] \times [0,1] \to B$ between the two ($H(t, 0) = g(t)$ and $H(t, 1) = h(t)$) that fixes the base point ($H(0,s) = H(1,s) = b_0$).

As you probably know, covering spaces also lift homotopies (this is proposition 1.30 in Hatcher's book), say $\hat{H} : [0,1]\times[0,1] \to E$ such that $p \circ \hat{H} = H$ and $\hat{H}(t,0) = \hat{g}(t)$. Since the fiber is discrete, $\hat{H}(0,s) = \hat{b}_0$ for all $s$ and therefore $\hat{H}(t,1) = \hat{h}(t)$ because the two lifts of $h$ agree at one point.

You therefore get a path $\gamma(s) = \hat{H}(1,s)$ between $\hat{g}(1)$ and $\hat{h}(1)$. But since the homotopy $H$ fixes the base point, the whole path lies in the fiber, which is discrete, and therefore it's constant and $\hat{g}(1) = \hat{h}(1)$.

Answer 2

Yes, that's right. If $f$ is a nullhomotopic loop in $B$ at $b_0$ and $e_0$ is any point in the fiber of $b_0$, then the homotopy rel $\partial I$ from $f$ to the constant map lifts to a unique homotopy rel $\partial I$ from $f$'s lift $\tilde f$ at $e_0$ to the lift of the constant loop, i.e. the constant path at $e_0$. This implies that $\tilde f$ is closed in $E$.