= Solution
Surjectivity was proved in part (b). Fix $q\in N$ and choose a normal ball $B=B_N(q,\varepsilon)$ on which $\exp_q$ is a <diffeomorphism> from the tangent ball. For each $p\in f^{-1}(q)$ define a section
$$
s_p(\exp_qv)=\exp_p\bigl((df_p)^{-1}v\bigr),\qquad |v|<\varepsilon.
$$
Completeness of $M$ guarantees that the source <Riemannian exponential map> is defined. Preservation of <geodesics> and uniqueness of their initial-value problem give $f\circ s_p=\mathrm{id}_B$. Since $df$ is invertible, differentiation of this identity shows that $s_p$ is a local <diffeomorphism>; it is injective because it is a section. Its image $U_p$ is open, and $f:U_p\to B$ is a <diffeomorphism>.
These images are disjoint. If $s_p(y)=s_{p'}(y)$, lift the unique radial <geodesic> from $y$ back to $q$ starting at that common point. Both section constructions give this same lift, and uniqueness forces its endpoint to be both $p$ and $p'$. Conversely, for any $x\in f^{-1}(B)$, lift that reversed radial <geodesic> from $f(x)$ to $q$ starting at $x$. Completeness supplies its full finite interval; its endpoint is some $p\in f^{-1}(q)$. Reversing the lift shows $x=s_p(f(x))$. Consequently
$$
\boxed{f^{-1}(B)=\coprod_{p\in f^{-1}(q)}U_p,\qquad f|_{U_p}:U_p\longrightarrow B\text{ is a diffeomorphism}.}
$$
Every point has such an evenly covered neighborhood, proving that \b[the complete local <isometry> is a covering]. This argument does not assume a uniform positive source injectivity radius.
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