Here is a quick question regarding expanding maps.
Suppose that $M$ is a closed (compact) manifold endowed with a Riemannian metric.
Take a map $F: M \to M$ which is continuously differentiable and satisfies
$$\inf_{\substack{\phantom{a}\\ x\,\in\,S}}\left(\inf_{\substack{\phantom{a}\\ v\,\in\, T_xM}} \left.\frac{\|D_xFv\|}{\|v\|} \right.\right)\ge \lambda > 1,$$
where the norms above are those of the Riemannian metric in the tangent spaces of $x$ and $F(x)$.
Question. Is this a sufficient condition to conclude that the number of preimages of any point is bounded across $M$?
i.e., Does it follow from these assumptions that
$$ \sup_{x\in S}\big|\, F^{-1}(x)\, \big|< \infty\;?$$
If not, in what settings/dimensions does it hold, or what further assumptions are required?
Okay, I think I have it, with the help of @JohnB's comment.
Indeed it is true that $F$ must be a local diffeomorphism (as the inf-inf equation means that $D_xF$ has no zero eigenvalues and therefore is of full rank, for any $x$) and that's enough to conclude.
Firstly, it follows that a given $x\in M$ can have only finitely many preimages: otherwise, by compactness of $M$, its set of preimages $F^{-1}(x)$ has an accumulation point; and there's no way $F$ can be injective on a neighbourhood of this point $\leadsto$ contradiction.
Secondly, the function $x \mapsto |F^{-1}(x)|$ is locally constant and therefore constant on each of the [finitely many] connected components of $M$. That affirms my question. This is a little harder to prove, but here is a sketch:
Suppose that $F^{-1}(x) = \{y_1,\ldots y_n\}$. We can certainly find/construct open neighbourhoods, $U,V_1,\ldots V_n$ of $x,y_1,\ldots y_n$ respectively, such that $F:V_i\to U$ is a diffeomorphism for each $i$.
It is immediate that every point in $U$ must have at least $n$ preimages, one in each of the $V_i$. Moreover, if there is a sequence of points in $U$, $(x_k)_{k=1}^\infty \to x$ each with more than $n$ preimages, we can create a sequence $(z_k)$ with $F(z_k) = x_k$ and the property that each $z_k$ does not lie in any $V_i$. Again by compactness, we can take $z$ as an accumulation point of this sequence, and by continuity it must lie in $F^{-1}(x)$, therefore one of the $V_i$'s.
Thereby $(z_k)$ would have to enter one of the $V_i$'s, a contradiction.