I was working through some things in riemannian geometry and I had this doubt:
Let $M$ be a closed riemannian manifold, $H$ an embedded submanifold and $V$ be its $\varepsilon$-tubular neighborhood. Consider a sequence of continuous curves $\sigma_{l}:[0,1] \rightarrow M\setminus V$. Suppose that the sequence converges uniformly to a continuous curve $\sigma:[0,1]\rightarrow M\setminus V$. Can I assume that $\sigma$ is homotopic to some $\sigma_l$ for $l$ large enough?
I was thinking a differently than Jeremy. My argument is not nearly as general, since it relies explicitly on the metric.
Let $\sigma_l\to\sigma$ uniformly in $M$, where $M$ is some Riemannian manifold (not necessarily complete) without boundary. Since the convergence is uniform, for every $\epsilon>0$ there is some $N$ so that $l>N$ implies $\sup_t d(\sigma_l(t),\sigma(t)) < \epsilon$.
Since $\sigma$ is compact, it admits a tubular neighborhood $\nu$ of radius $\epsilon_0 > 0$. (Compactness lets us avoids pathologies such as an infinite spiral toward the "edge" of an incomplete manifold.) Associated to $\epsilon_0$ is an $N_0$ so that for all $l>N_0$, $\sigma_l$ lies entirely within $\nu$.
The tubular neighborhood is diffeomorphic to the normal bundle of $\sigma$ in $M$, so we may regard $\sigma_l$ as a section of the normal bundle. All sections of a vector bundle are homotopic to its zero section Pulling this homotopy back by the diffeomorphism between the normal bundle and $\nu$, we have a homotopy between $\sigma_l$ and $\sigma$. The minimal $N_0$ so that all $l > N_0$ have $\sigma_l$ homotopic to $\sigma$ is that given by the radius of the cut locus of $\sigma$.
The intuition here is that since $\sigma$ is compact, some small $\epsilon$-neighborhood of $\sigma$ must retract onto it (this is the tubular neighborhood). Since the $\sigma_l\to\sigma$ uniformly, for large enough $l$ we can "suck" the $\sigma_l$ onto $\sigma$ via that retraction.