Let $f:X \rightarrow Y $ be covering map.Prove that $f$ is fibration.
Proof:
Let $Z$ be topological space, $g: Z \rightarrow X$ map and $F:Z \times I \rightarrow Y$ homotopy such that $F(z,0)=f(g(z))$ $\forall z \in Z$. We claim that there exist a homotopy $\tilde{F}:Z \times I \rightarrow X$ such that $\tilde{F}(z,0)=g(z)$ $\forall z \in Z$ and $f \circ\tilde{F}=F$.
For each $z \in Z$ $\alpha_z: I \rightarrow Y$ $\alpha_z(t)=F(z,t)$ is a path in $Y$ with beginning in $f(g(z))$ so there exist a unique lift $\tilde{\alpha}_z$ with beginning in $g(z)$.
We define $\tilde{F}(z,t)=\tilde{\alpha}_z(t)$
Now we have $f \circ \tilde{F}=F$ and $\tilde{F}(z,0)=g(z)$ .All that is left is to prove that $\tilde{F}$ is continuous.
Can someone help me to prove continuity?