# Smooth map with null differential at each point are constant on the connected component of the domain

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?

• Yes, this looks perfect to me. Commented Jan 13, 2019 at 0:33
• Thank-you prof. Shifrin :D Commented Jan 13, 2019 at 11:10
• How can you suppose $U$ is connected? Commented Oct 8, 2020 at 3:30

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.)
• @Ted Shifrin If I take an atlas for $M$, assuming it is connected, can I supposed that the convers $U$ are path connected, and so the associate $\varphi(U)$ is path connected? Because if not, then how would your hint be relevant? Commented Oct 8, 2020 at 3:55
• By "the chain rule is your friend" do you mean by defining $g(t):=F(tx+(1-t)p)$, $t \in [0,1]$ which can be done on any element of the atlas on $M$ as they're all path connected? and then differentiating and invoking chain rule we use that the derivative is zero. Thus $g'(t)=F'(tx+(1-t)p)(x-p)=0$ Commented Aug 5, 2022 at 18:26