I have a quick question. I am asked to prove that if $F : X \to Y$ is a smooth map between connected manifolds with zero jacobian, then it is constant.
I found a proof which I am quite satisfied with. Basically, we restrict our study to charts around $x$ and $F(x)$, and define the map induced by $F$ between these two charts (so from $\mathbb{R}^n$ to $\mathbb{R}^m$ for some $n, m$). It is straightforward to check that this induced map has zero derivative, because it takes every vector in $T_xX$ to the zero vector in $T_{F(x)}Y$. So then, we see that $F$ is constant on this given chart for $X$, and from the connected hypothesis of $X$, $F$ is always constant.
My question is… why do we need connectedness of $Y$? If, in any case, $F$ is going to be constant on any chart of $X$, I don't understand why it would matter that $Y$ is connected.
If any one has any insight on something I may have missed, I'd be grateful!
Thanks 😀
Best Answer
It is irrelevant whether $Y$ is connected. Since $F$ is locally constant on $X$, it is constant on every connected component of $X$. Thus we only need $X$ connected. Another way to see it is to observe that $F(X)$ is connected, thus contained in a component $Y_0$ of $Y$. But $Y_0$ is a connected manifold.