$f:M\to N$ is a smooth map between manifolds of dimensions $m\geq n$. If $y\in N$ is a regular value, then the set $f^{-1}(y)$ is a smooth manifold of dimension $m-n$ or $\emptyset$.
What if $m<n$? My intuition tells me that $f^{-1}(y)$ is still a zero dimensional manifold but I don't know how to prove this.
As Grumpy Parsnip commented, the differential cannot be surjective for dimension reasons, so no point of $M$ is a regular value of $f$. Note that $f^{-1}(y)$ can look quite surprising in general; for instance, it can have infinitely many components (just look at the real line wrapped around $S^1$ via the morphism $\mathbb R^1 \to S^1$ and take the fiber of a point, which gives something like $\mathbb Z$). However, this is a case where the differential is surjective, so it is in no means a pathological situation. I thought it might shed some light on your question since I have an example of two connected manifolds and a fiber which has lots of components.
Hope that helps,