Can the inverse image of a regular value be a single point?

Question

I've just started self-studying differential geometry, using "differential geometry of curves and surfaces" by Do Carmo. I stumbled into a problem and I can't seem to figure it out.

The author proves that, given a differentiable function $f: U\subset R^3 \rightarrow R$, the inverse image $f^{-1}\left(\alpha\right)$ of a regular value $\alpha \in f(U)$ is a regular surface. To do this, he defines a new function $F(x,y,z)=(x, y, f(x,y))$, picks a point $p\in f^{-1}\left(\alpha\right)$, and uses the inverse function theorem to show that there are neighbourhoods in $p$ and $F(p)$ in which F has an inverse, $F^{-1}(u,v,t) = (x, y, g(u, v,t))$. He then shows that $h(x,y)=g(u, v, \alpha)$ is is the graph of the inverse image $f^{-1}\left(\alpha\right)$. Based on a previous proposition, this proves that $f^{-1}\left(\alpha\right)$ is indeed a regular surface.

My question is the following: what prevents $f^{-1}\left(\alpha\right)$ from being a single point, or a set of discontinuous points (in which case I suppose it wouldn't be a regular surface)?

Thanks in advance for any help

Answer 1

Not an answer, but a sequence of thought-provoking questions

The author proves that the inverse image of a regular value of a function of a particular type. What type of function is $f$ in your description? What's its domain? What's its codomain? If $f(p) = \alpha$, but for all $q$ near $p$, we have $f(q) \ne \alpha$, does that tell you anything about the image of a small ball near $p$? What about the derivative of $f$ at $p$?

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post-comment continuation

You have $f: \Bbb R^3 \to \Bbb R$, and $f(p) = q$. So for any closed ball $K$ (of radius $r > 0$, around $p$, we have $f(K)$ is a compact connected subset of $R$, which is a closed interval $[s, b]$ that contains $q$. Suppose $a < q < b$, i.e., that $q$ is not one of the endpoints. There's some point $A \in K$ with $f(A) = a$, and a point $B \in K$ with $f(B) = b$, and $A \ne B$ (why?).

Consider two paths $\alpha, \beta$ from $A$ to $B$ within $K$ --- there are infinitely many --- that share no pointes except their endpoints. For any such path $\gamma: [0, 1] \to K$, we have $$ f(\gamma(0)) = f(A) = a\\ f(\gamma(1)) = f(B) = b. $$ so there's a value $t$ with $f(\gamma(t)) = q$. So there's a point $P_1$ on $\alpha$ with $f(P_1) = q$, and a different point $P_2$ on $\beta$ with $f(P_2) = q$. So the preimage of $q$ contains at least two points.

We've shown that if $q$ is not an endpoint of $[a,b]$, then $f^{-1}(q)$ is not a single point. So...$q = a$ or $q = b$. Let's look at the case $q = a$.

What's the derivative of $f$ at $p$ in some direction $v$? It's \begin{align} D_v f(p) &= \lim_{h \to 0} \frac{1}{h}(f(p + hv) -f(p))\\ &= \lim_{h \to 0} \frac{1}{h}(f(p + hv) - q)\\ &\ge \lim_{h \to 0} \frac{1}{h}(q - q)& \text{because $f$ only takes on values in [q, b]} \\ &= 0. \end{align}

So all directional derivatives are nonnegative, which means that they must all be zero (because the derivative in direction $v$ must be the negation of the derivative in direction $-v$). So the gradient of $f$ at $p$ is zero. So $q$ is not a regular value.

Answer 2

There's the local canonical form for submersions: $(x_1,\dots,x_n)\mapsto(x_1,\dots,x_m)$, where $n\ge m$ are the dimensions of the respective manifolds. (In your case I guess $n=3, m=1$.)

In particular, to get a surface we will need $n=m+2$. Thus we will not have $f^{-1}(\alpha)$ being a single point. It might be a disconnected set, though the connected components will be surfaces.