Relation between uniform convergence of curves in a manifold and homotopy 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?
 A: I assume you either want 1. a fixed endpoint homotopy of curves all having the same start point and the same end point or 2. a free or based homotopy of loops. Otherwise, it would be good to specify what kind of curves you have and what type of homotopy you want.
In either of these situations, the answer is yes. In fact, it holds in a much more general setting.
In a metric space $X$ which has a universal cover (locally path connected, semi-locally simply connected), if $\sigma_l\to\sigma$ uniformly, then there is a $N$ such that $\sigma_l$ is homotopic (in any of the above senses) to $\sigma$ for each $l\geq N$.
Really, you don't even need the metric. If $X$ is path connected, locally path connected, and semi-locally simply connected (the usual ingredients for a universal covering), and you use the compact-open topology on the space of paths (this agrees with the uniform metric when $X$ is metric), then if $\sigma_l\to \sigma$ in this path space, then there is a $N$ such that $\sigma_l$ is homotopic (in any of the above senses) to $\sigma$ for each $l\geq N$.
The general idea is, for large $l$, to connect $\sigma$ and $\sigma_l$ with "small" paths so that you get a sequence of "small" loops between the two. Since small loops in a space with a universal cover are trivial (homotopically), you can piece together all these homotopies to get a homotopy of $\sigma$ and $\sigma_l$.
There is a nice detailed proof of the general statement in 
Discreteness and Homogeneity of the Topological Fundamental Group
A: 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.
