Does the pullback metric preserve geodesic completeness?

Question

Let $\pi: M \to N$ be an immersion. If $g$ is a Riemannian metric on $N$, then it follows the pullback $\pi^*g$ is a Riemannian metric on $M$. Suppose $(N,g)$ is geodesically complete. Does it follow $(M,\pi^*g)$ is geodesically complete? My first thought is yes. If $\gamma(.)$ is a unit speed geodesic on $M$, then $\pi \circ \gamma$ is a unit-speed geodesic on $N$. Since $(N,g)$ is complete, then $\pi \circ \gamma$ extends forever to an infinite length geodesic. Though, I am unable to understand if I can translate that back to the original geodesic $\gamma$ also extending forever.

Answer 1

Consider the inclusion $i : \mathbb{R}^n\setminus\{0\} \to \mathbb{R}^n$. The Euclidean metric $g$ on $\mathbb{R}^n$ is geodesically complete, but $i^*g$ is not.