if the pushforward is zero on a connected manifold then the function is constant.

Question

Let $F:M \to N$ be a smooth map between manifolds, and suppose $M$ is connected and that

$$F_*: T_pM \to T_{F(p)}N$$

is zero for every $p \in M$. Then $F$ is the constant map.

Here is mt attempt:

Let $q \in N$ with $$F(p)=q.$$ I claim the set $$A=\{x \in M : F(x)=q\}$$ is both open and closed in $M$. Clearly it is nonempty as $p \in A$. And by connectedness of $M$, this would force $A=M$ thus $F$ is constant on $M$. For closed let $x_0 \in M$ and $\{x_n\} \subset A$ with $$\lim_{n \to \infty}x_n=x_0.$$ As $x_n \in A$ for every $n$, $F(x_n)=q$ and by continuity of $F$, we have that $$F(x_0)=q$$ forcing $x_0 \in A$ thus $A$ contains its limit points and is closed in $M$. To show $A \subset M$ is open, fix some $a \in A$ and let $\epsilon>0$ be given such that $$B_\epsilon(a) \subset M.$$ as $M$ is connected, it is path connected thus if $z \in B_\epsilon(a)$, then define $$g:[0,1] \to M$$ via $$g(t):=F(zt+(1-t)a)$$ so that $g(0)=F(a), g(1)=F(z).$ and $g$ is a path. Then $$g_*=F_*(zt+(1-t)a)(z-a)=0.$$ Thus $g_*(t)=0$ for $t \in [0,1]$ implying $g$ is constant thus $$F(z)=g(1)=g(0)=F(a).$$ I.e., $B_\epsilon(a) \subset A$ and $A \subset M$ is open thus is all of $M$ and $F$ is thus constant on $M$.

For an exmaple of when connecedness is dropped, then it fails, consider $$f: \Bbb{R} \setminus \{0\} \to \{-1,1\}$$ via $$f(x)=\begin{cases} 1 & x>0\\ -1 & x<0 \end{cases}.$$ This function has zero derivative but is non constant. Furthermore the domain is a disconnected manifold.

Added: I am a bit shaky on the showing $A$ is open in $M$ but certain my closed argument works just fine. If I am correct, an upvote or a "looks ok" comment will suffice, if I am incorrect somewhere point it out and give a hint as to how to fix it please, thanks in advance! Also I think my counterexample works just fine

Answer 1

If $F_{*}=0$ then its matrix representation is the zero matrix. Hence locally, the partial derivatives of $F$ vanish. Hence locally, $F$ is constant. Since the manifold is connected, $F$ is constant everywhere.

Answer 2

Suppose that there are two points in the image, and pick a function $f:N\to\mathbb R$ that takes different values at those two points. The function $f\circ F$ is then not constant, but the chain rule tells you that its differential is zero.

Answer 3

I think you can generalize the "$A$ is open"'s argument to prove the entire statement. First of all lets fix an $x_0\in M$ and define $F(x_0)=q$, now we have only to prove that $F(x)=q\;\; \forall x\in M$.

By $M's$ connectedness there exists some smooth path $\phi(t)$ from $x_0$ to $x$ (i'd avoid using $tx_0+(1-t)x$ because we're not allow to treat points like vectors in a generic smooth manifold), now the following path on $N$ $$\alpha(t):=F(\phi(t)),\;\; \alpha'(t)=D_{\phi(t)}F (\phi'(t))\stackrel{F_*=0}{=} 0 \;\; \forall t, $$ has zero speed, so for the The Existence and Uniqueness Theorem for Solutions to ODEs, $\alpha(t)$ must be the constant path hence $$F(x)=F(\phi(1))=\alpha(1)\equiv \alpha(0)=F(\phi(0))=F(x_0)=q. $$