Suppose $\Gamma : I=[0,1] \rightarrow X$ is a nullhomotopic in a topological space $X$. I want to show that any lift $\widetilde{\Gamma}:I \rightarrow \widetilde{X}$ in a covering space $\widetilde{X}$ must be a loop.
I want to verify that my proof is correct.
We can think of a loop a map from the circle with basepoint, i.e. $\Gamma:(S^1,s_o) \rightarrow \pi_1(X,x_0)$. The induced map $\Gamma_* : \pi_1(S^1,s_o) \rightarrow \pi_1(X,x_0)$ is trivial since $\Gamma_*[1]=[\Gamma] \in \pi_1(X)$, where I'm using $[1]$ to denote the class of the loop generating loop $1$ going exactly once around the circle. By the lifting criterion this then lifts to a map $\widetilde{\Gamma}: (S^1,s_0) \rightarrow (\widetilde{X},x_0)$ which is a loop in $\widetilde{X}$.
Best Answer
Your proof is correct. This community wiki solution is intended to clear the question from the unanswered queue.