For a Riemannian covering map $\tilde M\rightarrow M$, any metric on $\tilde M$ invariant under deck transformations descends $M$? This is a problem in Riemannian Manifolds: An Introduction to Curvature (GTM176), Chapter 3, Exercise 3.6. The problem states

If $\pi : \tilde M \to M$ is a smooth covering map, and $\tilde g$ is any metric on $\tilde M$ that is invariant under all covering transformations, show that there is a unique metric $g$ on $M$ such that $\tilde g = \pi ^* g$.

I know the first step is to constructing $g$ by the pull-back of $\pi^{-1}$ at local coordinates of every points, then show that this definition would not conflict for different points show that for any two points $\tilde p_1, \tilde p_2 \in \tilde M$ satisfying $\pi (\tilde p_1) = \pi (\tilde p_2) = p$.
My idea is: if I can select a covering transformation $\varphi: \tilde M \to \tilde M$, such that $\varphi(p_1) = \varphi(p_2)$, then I can use $\pi^* = \varphi^* \circ \pi^*$ to draw the conclusion that the pullback of $g|_p$ at $p_1$ and $p_2$ are exactly $\tilde g|_{\tilde p_1}$ and $\tilde g|_{\tilde p_2}$.
However, I realized that I don't really know how to prove that such a covering transformation $\varphi$ really exists! The topic of covering map and covering transformations (seems to relating to differential and algebraic topology too much) is too unfamiliar to me. So do anyone know how to prove:
If $\pi: \tilde M \to M$ is a covering map between smooth manifolds, then for every $\tilde p_1, \tilde p_2 \in \tilde M$ such that $\pi (\tilde p_1) = \pi (\tilde p_2) = p$, there exists a covering transformation (deck transformation) such that $\varphi(p_1) = p_2$?
 A: Your intuition is correct: covering transformations need not always exist when you want them to.
And, in fact, the statement in your post is true only for normal covering maps (also known as regular covering maps or Galois covering maps), meaning covering maps $\pi : \tilde M \to M$ having the property that for each $p \in M$, the deck transformation group of $\pi$ acts transitively on the fiber $\pi^{-1}(p)$.
For a covering map that is not normal, the statement is false. There exist examples of covering maps $\pi : \tilde M \to M$ that are not homeomorphisms for which the deck transformation group is trivial. Any Riemannian metric on $\tilde M$ is therefore invariant under all covering transformations, for the simple reason that the identity map is the only covering transformation. But it is far from true that any Riemannian metric on $\tilde M$ is the lift of a metric on $M$.

Here's one way you can corect the statement in your post to get a statement that is actually true.
Consider $U \subset M$ an open subset that is evenly covered by the map $\pi$, meaning that $\pi^{-1}(U)$ can be expressed as a union of pairwise disjoint collection $\{V_\alpha\}_{\alpha \in A}$ of open subsets $V_\alpha \subset \tilde M$, such that each restriction $\pi \mid V_\alpha : V_\alpha \to U$ is a diffeomorphism. For any $\alpha,\beta \in A$ there is a unique diffeomorphism $h:V_\alpha \to V_\beta$ such that $(\pi \mid V_\beta) \circ h = \pi \mid V_\alpha$. Let me refer to such a diffeomorphism $h$ as a local deck transformation over $U$.
The correct statement is that if $\tilde g$ is a metric on $\tilde M$, and if $\tilde g$ is invariant under every local deck transformation over every evenly covered open subset of $M$, then there is a unique metric $g$ on $M$ such that $\tilde g = \pi^* g$.
