Let $F:M\to N$ be a smooth map between smooth manifolds $M$ and $N$ (with or without boundary).
I want to show that $dF_p:T_pM\to T_{F(p)}N$ is the zero map for each $p\in M$ if and only if $F$ is constant on each component of $M$.
Here is my argument:
Suppose $F$ is constant on each component of $M$, and let's show that $dF_p:T_pM\to T_{F(p)}N$ is the zero map for each $p\in M$.
Let $p\in M$ and let $U$ be the component of $M$ containing $p$. Since $M$ is locally path connected, I know that $U$ is open in $M$. By hypothesis we have $F_{|U}:U\to N$ is consant. Then $d(F_{|U})_p:T_pU\to T_{F(p)}N$ is the zero map: let be $v\in T_pU$ and $f\in C^{\infty}(N)$, then $d(F_{|U})_p(v)(f)=v(f\circ F_{|U})=0 $ since $f\circ F_{|U}$ is costant from $U$ to $\mathbb{R}$. Since $d(\iota)_p:T_pU \to T_pM$ is isomprhism, and since we have that $dF_p\circ d(\iota)_p=d(F_{|U})_p$, we have that also $dF_p$ is the zero map.
We have to prove the converse (But here my problems begin) The best I have thought is this:
For simplicity suppose $M$ itself is connected. We know that $dF_p:T_pM\to T_{F(p)}N$ is the zero map for each $p∈M$ and we have to prove that $F:M→N $ is constant.
I want to show that $F$ is locally constant, i.e. each point $p$ in $M$ has an open neighborhood in $M$ such that $F$ is constant on this neighborhood.
Let $p\in M$ and let $(U,\phi=(x^1,\dots,x^m))$ be a chart on $M$ in $p$. Then $\{{\frac{\partial}{\partial x^i}}|_p\}$ is a basis for $T_pM$. By hypothesis we know that $dF_p(\frac{\partial}{\partial x^i}|_p)(f)=0$ for each $f∈C^∞(N)$, i.e. $\partial_i|_{\phi(p)}(f\circ F \circ \phi^{-1})=0$. We can suppose that $U$ and thus $\phi(U)$ are connceted, so by Ordinary Analysis we have thaht $f\circ F \circ \phi^{-1}$ is constant on $\phi(U)$. But $\phi$ is a diffeomorphism, so we have thaht $f \circ F:U\to N$ is constant, for each $f∈C^∞(N)$.
Now suppose there are $p\ne q \in M$ such that $F(p) \ne F(q)$. I want to construct a function $f∈C^∞(N)$ such that $f(F(p))\ne f(F(q))$.
The best I come out is: suppose there is a smooth chart $(V,\psi)$ on $N$ containing $F(p)$ and $F(q)$ and such that there is $K$ closed subset of $N$ such that $K\subseteq V$. Since $\psi $ is injective, then $\psi (F(p))\ne \psi( F(q))$, so they differ by at least a component, say the $j$ component. Let $\pi_j:\mathbb{R}^n\to \mathbb{R}$ the $j$ projection, and consider $\psi \circ \pi_j:\psi(V)\to \mathbb{R}$. Extend this function to a function $f∈C^∞(N)$ such that $f$ and $\psi \circ \pi_j$ agree on $K$. Then we have $f(F(p))\ne f(F(q))$ which is a contraddiction.
I know that this is quite completely wrong (I did many not-necessarily-true assumptions). And maybe this argument does not work in the case of manifolds with boundary.
So can anyone help me with observations/ sugestions/ hints, or even a full solution? Thank you.
I mention that this is Problem 3.1 in John Lee's Book "Introduction to smooth manifolds, 2 edition"
EDIT Thanks to the hint of @Ted Shifrin I come with another argument.
Let's start from a fact that I konw: if $A$ is an open subset of $\mathbb{R}^n$ and $A$ is connected, then each smooth function $f:A\to \mathbb{R}$ whose partial derivatives are zero in $A$, is constant.
Now we can generalize this as: if $A$ is an open subset of $\mathbb{R}^n$ and $A$ is connected, then each smooth function $f:A\to \mathbb{R}^m$ such that all component functions have partial derivatives that are zero in $A$, is constant. (We can deduce this from the former, simply noting that each component function is constant, right?)
Now, let $p\in M$ and $(U,\phi)$ smooth chart on $M$ in $p$ and $(V,\psi)$ smooth chart on $N$ in $F(p)$ with $F(U)\subseteq V$. Then $\psi \circ F \circ \phi^{-1}:\phi(U)\to \psi (V)$ is smooth, and we can suppose $U$ is connceted, and so is $\phi(U)$.
We have that $d(\psi \circ F \circ \phi^{-1})_{\phi(q)}=d\psi_{F(q)} \circ dF_q \circ d(\phi^{-1})_{\phi(q)}$ and since $dF_q$ is zero for all $q \in U$, then also $d(\psi \circ F \circ \phi^{-1})_{x}$ is zero for all $x \in \phi(U)$. But this is the jacobian matrix of $\psi \circ F \circ \phi^{-1}:\phi(U)\to \psi (V)$ . So by the above discussion we have that $\psi \circ F \circ \phi^{-1}:\phi(U)\to \psi (V)$ is constant, and then $F$ is constant on $U$, right?
Then, since $F$ is locally constant, we have that $F$ is constant on each component of $M$, right?
Is my new argument correct? Have I used well the Ted's hint?
Best Answer
HINT: Show that a smooth (vector-valued) function with zero derivative on a path-connected open subset of $\Bbb R^n$ is constant. (The chain rule might be your friend.)