Suppose $M$ and $N$ are smooth manifolds without boundary. I'd like to show that $F:M\to N$ is both a smooth immersion and a smooth submersion whenever $F$ is a local diffeomorphism. Fix $p\in M$. By employing the definition of a local diffeomorphism, John M. Lee, the author of my textbook, claims that there exists a neighborhood $U$ of $p$ s.t. $F$ maps $U$ diffeomorphically onto $F(U)$. From this, he concludes that $dF_p:T_p M\to T_{F(p)}N$ is an isomorphism. This confuses me a lot. As far as I know, one can say so when $F$ is globally a diffeomorphism. But, in our case, $F$ is merely a local diffeomorphism. What's the argument for this, please? BTW, I haven't come to the chapter about submanifolds yet. Thank you.
$F:M\to N$ is a local diffeomorphism iff it is both a smooth immersion and a smooth submersion.
differential-geometrysmooth-manifoldssubmanifold
Related Solutions
- The notion of a local diffeomorphism makes sense only if the domain and the range are smooth manifolds. If the image of a map happens to be a smooth submanifold of the target manifold, one can say " $f$ is a local diffeomorphism onto its image" by restricting the range. Any other use is just made-up (by various MSE users, it seems) and should be avoided (at least until you are very comfortable with the subject). Instead you can simply say:
...The image of a map $f: X\to Y$ is a smooth submanifold and $f: X\to f(X)$ is a local diffeomorphism.
You may also sometimes encounter the following, describing what an immersion is:
A map $f: X\to Y$ of smooth manifolds is an immersion if and only if locally, it is a diffeomorphism to its image, meaning that $\forall x\in X \exists$ a neighborhood $U$ of $x$ such that $f(U)$ is a smooth submanifold of $Y$ and $f: U\to f(U)$ is a diffeomorphism.
But, again, given ambiguity of the language, it is better to avoid using this terminology in the beginning. The ambiguity comes from the word "image": It can either mean the image of the original map or the image of the map with the restricted domain.
- Everything that you wrote up to the line "However, the first and third posts..." is correct and proofs are very straightforward.
However: I did not check your guesses on how it may or may not be related to various MSE posts.
One thing, you should not repeat ad nauseum "with dimension". (Every manifold has dimension and, except for the empty set, its dimension as a smooth manifold equals its dimension as a topological space. As for the empty set: For every $n\ge 0$, the empty set is a manifold of dimension $n$. At the same time, from the general topology viewpoint, the empty set has dimension $-1$.)
- As for what various MSE users meant in their answers and comments, I prefer not to discuss: Frequently, there is no consistency in their use of mathematical terminology. (Many are only beginners, many have trouble with English, etc.)
Addendum. I am not sure who came up with the idea of allowing manifolds to have variable dimension on different connected components, but I wish this never happened as this just leads to a confusion. I checked several sources in geometry and topology, and the only author allowing manifolds to have variable dimension is Lang.
I think it does not work. The problem is that F surjective does not implies in general $ \hat{F}$ to being surjective. You are considering open sets when you are taking $\hat{F}$ and could happens that if you take $y$ in the codomain of $\hat{F}$, then there exists $x$ such that $F(x)=y$ but $x$ does not belong to the domain of $\hat{F}$.
I’m pretty sure that the main idea of the proof is that one that you explained, but you should fix this problem firstly.
Best Answer
This is one pedantic way to see it. Let $F(p) = q$ and $V=F(U)$. Suppose that $f: U \to V$ is the diffeomorphism obtained from the restriction of $F$ to open subsets $U \subseteq M$ and $V \subseteq N$.
We know that $df_p : T_pU \to T_{q}V$ is an isomorphism. We can relate $df_p$ to $dF_p$ through isomorphisms $di_p : T_pU \to T_pM$ and $dj_{q} : T_{q}V \to T_qN$, where $i : U \hookrightarrow M$ and $j : V \hookrightarrow N$ are the inclusion maps. We observe that $$\require{AMScd} \begin{CD} U @>{f}>> V\\ @V{i}VV @VV{j}V \\ M @>{F}>> N \end{CD}, \quad \text{that is} \quad j \circ f = F \circ i.$$ So the differentials related as $dj_q \circ df_p = dF_p \circ di_p$. All three of them known to be isomorphisms, so $dF_p$ is also an isomorphism. Another way is to compute directly the Jacobian matrix of the differentials on a chart, which is somewhat simpler.