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”