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Problem 3-1: Let $M$ and $N$ be smooth manifolds with or without boundary, and let $F:M\to N$ be a smooth map. Show that $dF_p:T_pM\to T_{F(p)}N$ is the zero map if and only if $F$ is constant on each component of $M$.

This is my attempt:

In local coordinates, we have $$dF_p\left(\frac{\partial}{\partial x^i}\Big|_p\right)=\frac{\partial F^j}{\partial x^i}(p)\frac{\partial}{\partial x^j}\Big|_{F(p)}\,.$$ Hence, $dF_p=0$ if and only if, for all $i$, $dF_p\left(\frac{\partial}{\partial x^i}|_p\right)=0$, if and only if $\frac{\partial F^j}{\partial x^i}(p)=0$ for all $i,j$, if and only if $F^j=$ constant for all $j$.

But after looking at this question I can see I am wrong and the solution is much more complicated.

Why is my answer wrong?

A. Goodier
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    my first remark is that there should be a "for all p" in the problem statement otherwise it is clearly false. Next, it might be useful to call the coordinates on the domain and target by different names, say $x $ on the domain and $y$ on the target space. I suggest you be more careful with your if and only ifs and state clearly where you use the conditions " for all $p$, $dF_p = 0$" and deduce carefully what this implies (because this condition does NOT imply the functions are globally constant) – peek-a-boo Jan 03 '20 at 12:27
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    @peek-a-boo is right: the problem is incorrectly quoted from my book. The second sentence actually says "Show that $dF_p\colon T_pM\to T_{F(p)}N$ is the zero map for each $p\in M$ if and only if $F$ is constant on each component of $M$." – Jack Lee Jan 05 '20 at 23:22

1 Answers1

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You seem to be jumping from the fact that all derivatives of $F$ at one point $p$ are $0$ to the conclusion that $F$ is constant. That is false, of course.

Even if you prove that all derivatives of $F$ at all points are $0$, then, when you start to work in local coordinates, you are working in with a chart $\psi\colon\mathbb R^n\longrightarrow M$. But then all that you shall have proved would be that the restriction of $F$ to $\psi(\mathbb R^n)$ is constant. You still will not have proved that the restriction of $F$ to each connected component of $M$ is constant.

José Carlos Santos
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