On the nature of a submanifold

Question

In the book "Topology from the Differential Viewpoint" (Milnor) on page 11 we get the following lemma:

If $f: M\to N$ is a smooth map between manifolds of dimension $m\geq n$ and if $y\in N$ is a regular value, then the set $f^{-1}(y) \subset M$ is a smooth manifold of dimension $m-n$.

My question is:

What are the conditions on the function $f$ to get:

a) $f^{-1}(y)$ is connected

b) $f^{-1}(y)$ is simply connected or Non-simply connected

c) $f^{-1}(y)$ is finite

d) $f^{-1}(y)$ is infinite

For example we know that $S²$ is simply connected and related to a well known function.

Answer 1

An answer for $(c)$:

Let $M$ and $N$ be smooth manifolds (of the same dimension), with $M$ compact, and let $f : M \to N$ be a smooth map, with $y \in N$ a regular value for $f$, then $f^{-1}(y)$ is a finite set.

Proof: Since $\{y\} \subseteq N$ is closed, and $f$ is smooth (and hence continuous), it follows that $f^{-1}(y) = f^{-1}[\{y\}]$ is closed in $M$.

Now suppose $f^{-1}(y)$ was not finite (and hence infinite), then $f^{-1}(y)$ would contain a limit point, say $p \in f^{-1}(y)$ since $M$ is compact (because infinite subsets of compact sets contains a limit point). The fact that $p$ is contained in $f^{-1}(y)$ follows from the fact that $f^{-1}(y)$ is closed in $M$. Since $p$ is a limit point of $f^{-1}(y)$, by definition every neighbourhood of $p$ contains at least one point, say $q$, in $f^{-1}(y)$ other than itself.

Now $f(q) = y$, which contradicts the inverse function theorem which says that $f$ is one-to-one in some neighbourhood $U'$ of each $x \in f^{-1}(y)$. Hence $f^{-1}(y)$ must be finite. $\square$

Edit: Look at page 8 of Milnor's book, which is where he claims this.

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Answer for $(d)$

If $f : M \to N$ is a smooth map between manifolds of dimension $m > n$, and if $y \in N$ is a regular value, then $f^{-1}(y)$ is infinite

Proof: Suppose $f^{-1}(y)$ was finite, then every neighbourhood $U \subseteq M \cap f^{-1}(y)$ of $x \in f^{-1}(y)$ contains only finitely many points. But since $f^{-1}(y)$ is a smooth manifold of dimension $m -n \geq 1$, some neighbourhood $V$ of $x$ must be homeomorphic to an open subset of $\mathbb{R}^{m-n}$. But any open subset of $\mathbb{R}^{m-n \geq 1}$ contains infinitely many points and since homeomorphims are bijective, we have a bijection between a finite set (namely $V$) and an infinite set, a direct contradiction. $ \square$

Comment: @BalarkaSen pointed this out to me, so credit should go to him for this part of the answer as well.

Answer 2

As others said, in general you cannot claim anything about a) and b) for a random $f$.

However, for c nd d we have this: When $n=m$, by being a regular point we can show that the set $f^{-1}\{y\}$ cannot have accumulation point. In other words it is a discrete set. Because of this, if the domain is bounded, then this set must be finite. If the domain is unbounded then this set can be either finite or countably large.