Isometries of Riemannian manifolds

Let $F : M \to N$ be a local isometry between connected Riemannian manifolds $M$ and $N$. Show that if $M$ is complete, so is $N$ and $F$ is a covering map.

Let $F : M \to M$ be an isometry of a Riemannian manifold $M$.

The space $\mathbb {R}^{n\times n}$ of $n\times n$ real matrices forms an $n^2$-dimensional Euclidean space, in which the dot product between $A=[a_{ij}]$ and $B=[b_{ij}]$ is given by

Let $S^{n-1}$ be the unit sphere in $\mathbb {R}^{n}$, and define $\sigma : S^{n-1} \to \mathbb {R}^{n\times n}$ as the map sending $\mathbf{x} = (x_1,\ldots ,x_n)$ in $S^{n-1}$ to the symmetric matrix $\sigma (\mathbf{x}) = \frac{1}{\sqrt2}[x_ix_j]$.

Let $\phi : \mathbb {R}^{n} \to \mathbb {R}^{n}$ be any isometry, that is, a map such that the Euclidean distance between any two points $x,y \in \mathbb {R}^{n}$ is equal to the distance between their images $\phi (x), \phi (y)$. Show that $\phi$ is affine linear, that is, there exists a vector $b \in \mathbb {R}^{n}$ and an orthogonal matrix $A \in O(n)$ such that for all $x \in \mathbb {R}^{n}$, $\phi (x) = Ax+b$.