Consider a compact Riemannian manifold $(M,g)$. Does there exist a small enough $\varepsilon$ such that for any maps $f$ and $g$ from $M$ to $M$, if $\sup_{x \in M} d(f(x), g(x)) < \varepsilon$, then $f$ and $g$ must necessarily be homotopic to each other.
In other words, given a compact Riemannian manifold $M$, is there a distance small enough such that functions from $M$ to $M$ that are within that small distance are actually the same in homotopy?
If anyone has any idea, or possibly a reference where this sort of question might have a (partial) answer, I'd be quite grateful.
There's such $\epsilon$: As suggested in the comment, one can find $\epsilon>0$ small so that if $x, y\in M$ and $d(x, y)<\epsilon$, there's an unique shortest geodesic joining $x, y$. Now for all $x\in M$, since $$ d(f(x), g(x)) <\epsilon,$$ there is an unique $v_x\in T_{f(x)} M$ so that $\exp_{f(x)}(v_x)= g(x)$. Thus the homotopy
$$ f_t(x) = \exp_{f(x)} (t v_x)$$
is a smooth homotopy between $f = f_0$ and $g = f_1$, provided we know that $x\mapsto v_x$ is smooth.
This is indeed the case: the map
$$ \Phi: TM \to M\times M, \ \ \ (x, v)\mapsto (x, \exp_x v)$$
is smooth and has invertible differential at the zero section: indeed we have $ d\Phi_{(x,0)} (v, w) = (v, w)$. Then there is a diffeomorphism between $U$ an open neighborhood of the zero section and $V \subset M\times M$ (an open neighborhood of the diagonal in $M\times M$). In particular, we can take (by compactness) $U$ to be of the form $\{(x, v) : \| v\|<\epsilon\}$. Thus we have $$v_x = \Phi^{-1} (f(x), g(x))$$ and $v_x$ is smooth.