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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?Screenshot of example

BCLC
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Jason
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  • what page is it on? – Hank Igoe Jan 08 '21 at 05:32
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    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 $p \in U$. – Kelvin Lois Jan 08 '21 at 05:33
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    The definition of submersion says that the DIFFERENTIAL is onto (not the map itself). – Nick Jan 08 '21 at 05:42
  • @HankIgoe it is at page no. 96 under section 8.8 of 2nd edition. – Jason Jan 08 '21 at 06:14
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    yeah both SiKucing and Nick are correct. the map $i$ itself need not be onto as submersion was defined as "the differential is onto". – Jason Jan 08 '21 at 06:21
  • @Nick Can you explain how $i_{,p}$ is surjective? I mean since $i_{,p}$ is the map $\iota_{}:T_pU\rightarrow T_pM$ and the domain and range of $i_{,p}$ are different which means there are elements in $T_p M$ that have no preimage in $U$. – Jason Feb 08 '21 at 07:11
  • @Nick how do you know it's surjective? Please see here – BCLC Apr 30 '21 at 09:50
  • @BCLC In the example of the inclusion map of an open subset $U \subset M$, if $p \in U$, then $T_pU = T_pM$, so the differential $i_*$ is the identity map on $T_pM$. – Nick May 06 '21 at 04:44
  • @Nick But $T_pU$ and $T_pM$ are not literally equal right? See the comments in the answer below as well as the link i sent you – BCLC May 06 '21 at 15:52
  • Sure, but they are isomorphic in an obvious way. – Nick May 12 '21 at 18:13

1 Answers1

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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.

M Khaled Bin-Lateef
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  • 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? – Jason Jan 11 '21 at 08:42
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    Because $\iota_{*}$ is the identity map – M Khaled Bin-Lateef Jan 11 '21 at 10:00
  • Shouldn't $\iota_{}:T_pM\rightarrow T_pM$ be $\iota_{}:T_pU\rightarrow T_pM$? Since the domain of $\iota$ is restricted to only $U$. – Jason Feb 08 '21 at 06:51
  • And now when the domain and codomain of $\iota_{*}$ are not equal how can this be a identity map? – Jason Feb 08 '21 at 07:04
  • @JasonLiam $T_pU=T_pM$ for all $p\in U$. – M Khaled Bin-Lateef Feb 08 '21 at 17:12
  • Do we have a proof for $T_p U=T_p M$. For example $U$ is a subset of $M$ and how can the tangent space at all points $p\in M$ be equal to tangent space at the same point in $U$? Shouldn't this be $T_p U\simeq T_p M$ instead of the strict equality. You only considered for $p\in U$ but didn't considered the converse. – Jason Feb 09 '21 at 05:45
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    If $U$ is open in $M$ then $C_p^{\infty}(U)=C_p^{\infty}(M)$ for all $p\in U$. – M Khaled Bin-Lateef Feb 10 '21 at 15:23
  • Moe, are you sure you can that $T_pU$ and $T_pM$ are literally equal and not just isomorphic? Same with the set of germs $C_p^{\infty} U$ and $C_p^{\infty} M$. In the sense that they are literally equal, of course it's easy to say $\iota$ is immersion and submersion: The differential is the identity! See here please – BCLC Apr 30 '21 at 09:52
  • If $C_{p}^{\infty}U=C_{p}^{\infty}M$ then $T_pU=T_pM$. Sure, if you want to be really technical and pedantic, then the germs are canonically isomorphic and its common practice to not distinguish between them. But yes, if you assert that $C_{p}^{\infty}(U)$ and $C_{p}^{\infty}(M)$ are literally equal, then $T_pU$ and $T_pM$ must be literally equal for $p\in U$. – M Khaled Bin-Lateef May 01 '21 at 11:55
  • In particular, if $C_{p}^{\infty}(U)=C_{p}^{\infty}(M)$ then $T_pU={ v_p:C_{p}^{\infty}(U) \rightarrow \mathbb{R}$ $:$ $v_p$ linear and satisfies product rule $}$ $=$ ${ v_p:C_{p}^{\infty}(M) \rightarrow \mathbb{R}$ $:$ $v_p$ linear and satisfies product rule $}$ $=T_pM$ – M Khaled Bin-Lateef May 01 '21 at 12:07
  • oh thanks Moe, though i wasn't notified because you didn't tag me – BCLC May 05 '21 at 07:32
  • what do you say to this and this too in addition to my answer in previous link please. – BCLC May 05 '21 at 07:33