I am reading "An introduction to Manifolds" by Loring Tu. There they've first defined immersion and submersion between manifolds and then gave an example. In the last line of the example they've written This example shows in particular that a submersion need not be onto. And i am unable to understand that statement. For reference the following is the way they've defined immersion and submersion:
A $C^\infty$ map $F:N\to M$ is said to be an immersion at $p\in N$ if its differential $F_{*,p}:T_pN\to T_{F(p)M}$ is injective and a submersion at p if $F_{*,p}$ is surjective.We call F an immersion if it is an immersion at every $p\in N$ and a submersion if it is a submersion at every $p\in N$. I am attaching the screenshot of the example they've given and there i have highlighted the statement that i could not understand. My doubt is that surjective by definition means onto so how can a submersion need not be onto?
$\begingroup$
$\endgroup$
10
-
$\begingroup$ what page is it on? $\endgroup$– Hank IgoeCommented Jan 8, 2021 at 5:32
-
3$\begingroup$ The map $i : U \to M$ is an inclusion map, where $U \subset M$ is open. So $i$ is injective but not onto, whereas $i_{*,p}$ is bijective for every $p \in U$. $\endgroup$– Kelvin LoisCommented Jan 8, 2021 at 5:33
-
4$\begingroup$ The definition of submersion says that the DIFFERENTIAL is onto (not the map itself). $\endgroup$– NickCommented Jan 8, 2021 at 5:42
-
$\begingroup$ @HankIgoe it is at page no. 96 under section 8.8 of 2nd edition. $\endgroup$– JasonCommented Jan 8, 2021 at 6:14
-
1$\begingroup$ yeah both SiKucing and Nick are correct. the map $i$ itself need not be onto as submersion was defined as "the differential is onto". $\endgroup$– JasonCommented Jan 8, 2021 at 6:21
|
Show 5 more comments
1 Answer
$\begingroup$
$\endgroup$
12
If $\iota: U\hookrightarrow M$ denotes the inclusion, where $U\neq M$ is open in $M$, then $\iota_{*}:T_pM\rightarrow T_pM$ is both, injective and surjective. Hence $\iota$ is an immersion and a submersion. But $\iota$ clearly need not be surjective.
answered Jan 9, 2021 at 2:53
-
$\begingroup$ Can you explain if $\iota: U\hookrightarrow M$ denotes the inclusion then why automatically $\iota_{*}:T_pM\rightarrow T_pM$ is both injective and surjective? $\endgroup$– JasonCommented Jan 11, 2021 at 8:42
-
4$\begingroup$ Because $\iota_{*}$ is the identity map $\endgroup$ Commented Jan 11, 2021 at 10:00
-
$\begingroup$ Shouldn't $\iota_{*}:T_pM\rightarrow T_pM$ be $\iota_{*}:T_pU\rightarrow T_pM$? Since the domain of $\iota$ is restricted to only $U$. $\endgroup$– JasonCommented Feb 8, 2021 at 6:51
-
$\begingroup$ And now when the domain and codomain of $\iota_{*}$ are not equal how can this be a identity map? $\endgroup$– JasonCommented Feb 8, 2021 at 7:04
-
2$\begingroup$ If $U$ is open in $M$ then $C_p^{\infty}(U)=C_p^{\infty}(M)$ for all $p\in U$. $\endgroup$ Commented Feb 10, 2021 at 15:23