Are the lifts of freely homotopic closed paths (with different start/endpoints) also closed freely homotopic paths? Weaker versions of this question have been asked before, but I am wondering about some more general cases:
Suppose that $\pi: E \to B$ is a covering space and $\alpha, \beta: [0, 1] \to B$ are closed freely homotopic paths (but we don't know if $\alpha(0) = \beta(0)$ or if $\alpha(1) = \beta(1)$). Assuming there is a closed lifting $\tilde{\beta}$ of $\beta$, must $\alpha$ have a closed lifting $\tilde{\alpha}$ as well? Must $\tilde{\alpha}$ be homotopic to $\tilde{\beta}$ in that case? Also, can we guarantee the existence of a closed lifting of $\beta$? If not, then must the liftings still be freely homotopic?
 A: Let's think of $\alpha$ and $\beta$ as being parameterized by the circle $S^1$:
$$\alpha : S^1 \to B
$$
$$\beta : S^1 \to B
$$
Now let's write out the assumptions:

*

*There exists a continuous function
$$H : S^1 \times [0,1] \to B
$$
such that $H(x,0)=\alpha(x)$ and $H(x,1)=\beta(x)$,

*There exists a lift $\tilde \beta : S^1 \to E$ of $\beta$. It follows that $\beta = \pi \circ \tilde\beta$.

I'll prove that there exists a lift $\widetilde H : S^1 \times [0,1] \to E$ of $H$ such that $\widetilde H(x,1)=\tilde \beta(x)$. It follows that $\tilde\alpha(x) = \widetilde H(x,0)$ is a lift of $\alpha$ and that $\widetilde H$ is a free homotopy from $\tilde\alpha$ to $\tilde\beta$.
The proof will use the general lifting lemma. Denote the base point on $S^1$ as $x_0$, denote the base point of $S^1 \times [0,1]$ as $(x_0,1)$, and denote $p = \beta(x_0) = H (x_0,1) \in B$. Also denote $\tilde p = \tilde \beta(x_0)$. To apply the general lifting lemma we need to check whether the image of the induced homomorphism
$$H_* : \pi_1(S^1 \times[0,1],(x_0,1)) \to \pi_1(B,p)
$$
is contained in the image of the induced homomorphism
$$\pi_* : \pi_1(E,\tilde p) \to \pi_1(B,p)
$$
We'll use the induced homomorphisms
$$\beta_* : \pi_1(S^1,x_0) \to \pi_1(B,p) \quad\text{and}\quad \tilde\beta_* : \pi_1(S^1,x_0) \to \pi_1(E,\tilde p)
$$
It follows that
$$\beta_* = \pi_* \circ \tilde\beta_*
$$
Since $S^1 \times [0,1]$ deformation retracts to $S^1 \times \{1\}$, it follows that
$$\text{image}(H_*) = \text{image}(\beta_*) = \text{image}(\pi_* \circ \tilde\beta_*) \subset \pi_*(\text{image}(\tilde\beta_*)) \subset \text{image}(\pi_*)
$$
and we're done.
