Why universal covering locally looks like the product I was reading Hatcher's Algebraic topology book,in page 69 the representing covering space using permutations section.
Use one of the fact that: Given the universal covering: $\tilde{X}_0 \to X$ for some nice space $X$, there is a neighborhood of $U \subset X$ over which the universal convering $\tilde{X}_0$ is $U\times \pi_1(X,x_0)$ (see bottom of page 69.)
This seems to be the local trivialization of the universal covering,In the construction of universal covering we consider the set $$\tilde{X}_0 = \{[\gamma]\mid \gamma(0) = x_0\}$$
Why locally $\tilde{X}_0$ looks like product $U\times \pi_1(X,x_0)$?`
 A: In fact the covering spaces $p : \tilde X \to X$ over a connected base $X$ are nothing else than the fiber bundles over $X$ with discrete fiber $F$. Discrete fiber essentially means that $F$ is nothing more a set which is topologized in a trivial way to regard the products $U \times F$, where $U \subset X$ is an evenly covered open subset, as topological spaces which are homeomorphic to the subspaces $p^{-1}(U) \subset \tilde X$ via a fiber-preserving $h : p^{-1}(U) \stackrel{\approx}{\to} U \times F$.
Thus your question is:

If $p : \tilde X \to X$ is the universal covering, why is $F = \pi_1(X,x_0)$?

Note that the group structure of $F$ is irrelevant as long as are we only interested in the fiber bundle structure of $p$.
Now recall Hatcher's Proposition 1.32:

The number of sheets of a covering space $p : (\tilde X, \tilde x_0) \to (X,x_0)$ with $X$ and $\tilde X$ path-connected equals the index of $p_∗(\pi_1(\tilde X, \tilde x_0))$ in $\pi_1(X, x_0)$ [which is the number of left (or right) cosets of $p_∗(\pi_1(\tilde X, \tilde x_0))$ in $\pi_1(X, x_0)$].

In other words, the fiber $F$ of $p$ can be taken as the set of left cosets of $p_∗(\pi_1(\tilde X, \tilde x_0))$ in $\pi_1(X, x_0)$.
In the universal covering we have $\pi_1(\tilde X, \tilde x_0) = 0$; thus $F = \pi_1(X, x_0)/0 = \pi_1(X, x_0)$.
