Hi i am trying to figure out why an inclusion map is both and immersion and submersion. This is what i have tried so far.Let $S$ be an open subset of a manifold $M$. Now the inclusion mapping is $\iota:S\to M$. Now to prove immersion/submersion we have to show that each of its differentials $\iota_{*,p}$, for $p \in S$, is, resp, injective/surjective. By definition of differential of a smooth map $F$,we have $(F_{*,p}(X_p))f=X_p(f\circ F)$ where $F_{*,p}:T_pM\to T_{F(p)}N$ where $M$ and $N$ are manifolds.Now, taking $F$ as the inclusion map $\iota$, we get $(\iota_{*,p}(X_p))f=X_p(f\circ \iota)=X_p(f)$ where $\iota_{*,p}:T_pS\to T_{\iota(p)=p}M$.

My questions are as follows:

How do we know from this that each $\iota_{*,p}$ is injective and surjective so that $\iota$ is both an immersion and a submersion?

  • $\begingroup$ What did you try to prove that $\iota_*$ is injective ? $\endgroup$ Feb 8 at 11:16
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    $\begingroup$ @PierreElis It is mentioned in the book "An introduction to Manifolds" by Loring at page 96 under section 8.8 that $\iota:S\to M$ is both immersion and submersion. And i am trying to prove it . $\endgroup$
    – Jason Liam
    Feb 8 at 11:48
  • $\begingroup$ I understand that. I am asking you about your ideas to prove this fact, because giving you the solution wouldn't really help you : you must try to do it yourself and unravel all the definitions. $\endgroup$ Feb 8 at 13:32
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    $\begingroup$ I don't have the answer that is why i am asking. I need the answer, it's okay if you can't provide it. $\endgroup$
    – Jason Liam
    Feb 8 at 15:02
  • $\begingroup$ @PierreElis i posted an answer. is it correct please? OP did in fact show ideas by giving the definition of the differential. I was actually stuck myself trying to answer and then I realised 'hey let's go back to the definition of the differential'. as it turns out this seemingly simple fact in section 8 may actually require facts from section 11 or section 13, at least if you want to be really precise. If you can accept that $T_pS=T_pM$ as a literal equality, then this question is actually pretty easy. But I don't think we can talk about those 2 vector spaces as literally equal. $\endgroup$
    – BCLC
    Apr 30 at 9:58

(I assume $M$ is a smooth $m$-manifold.)

The differential at $p \in S$ for $\iota: S \to M$ is $\iota_{*,p}:T_pS \to T_pM$. This differential is an $\mathbb R$-isomorphism of $\mathbb R$-vector spaces, i.e. it is bijective $\mathbb R$-linear. The key thing here is the relationship of $T_pS$ and $T_pM$ given that $S$ is open in $M$:

$T_pS \cong T_pM$ s.t. $\dim T_pS = \dim T_pM = m$

Proof surjective:

  1. $image(\iota_{*,p}) = \iota_{*,p}(T_pS)$ equals $range(\iota_{*,p}) = T_pM$ itself if $\iota_{*,p}(T_pS)$ is an $\mathbb R$-subspace of $T_pM$ and has the same dimension as an $\mathbb R$-vector space as $T_pM$.

  2. The image $\iota_{*,p}(T_pS)$ is an $\mathbb R$-subspace of $T_pM$: Because the image of an $\mathbb R$-linear map is an $\mathbb R$-subspace of the range of the $\mathbb R$-linear map.

  3. From rank-nullity theorem, $\dim \iota_{*,p}(T_pS) = \dim T_pS - \dim \ker(\iota_{*,p}) = m - \dim \ker(\iota_{*,p})$

  4. $\dim \ker(\iota_{*,p})=0$ iff $\iota_{*,p}$ is injective.

  5. $\iota_{*,p}$ is injective. (I'll show next.)

  6. By (4),(5) and (3), $\dim \iota_{*,p}(T_pS) = m$

  7. Therefore, by (6),(1) and (2), $image(\iota_{*,p}) = range(\iota_{*,p})$. QED surjective.

(Alternatively, perhaps you can say if they're the same dimension then 'injective if and only if surjective' or something. I've seen such statements for when domain and range are literally the same finite-dimensional vector space, but I haven't yet found such statement for merely isomorphic finite-dimensional vector spaces.)

Proof injective:


Wow this is tough. I notice you asked a similar question here a month ago referring to Tu's An Introduction to Manifolds. Actually it's in Section 8 that Tu says $i$ is an immersion. However, Tu doesn't quite prove $i$ is an immersion until Section 11, at least for regular $k$-submanifolds where $k$ need not equal $m$. Maybe the proof is a lot simpler for $k=m$. I just know that we can't exactly say that $T_pS = T_pM$, as if the underlying sets are literally equal. But of course in the sense that they are equal, $\iota_{*,p}$ is just the identity map, so this whole thing is pretty easy.

Let's not think that they are literally equal: Hmmm...

Let $\iota_{*,p}(X_p)=\iota_{*,p}(Y_p)$. Show $X_p=Y_p$, i.e. $X_pf = Y_pf$ for all $f \in C^{\infty}_{p}(S)$.

Let $g \in C^{\infty}_{p}(M)$. Then $\iota_{*,p}(X_p)=\iota_{*,p}(Y_p)$ implies $X_p(g \circ \iota) = \iota_{*,p}(X_p)(g)=\iota_{*,p}(Y_p)(g) = Y_p(g \circ \iota)$. Now, $g \circ \iota = g|_U$.

Sooo...if every $f \in C^{\infty}_{p}(S)$ (btw I've been treating a function $f \in C^{\infty}(S)$ the same as germ $[f] \in C^{\infty}_{p}(S)$, which is strictly incorrect) is the restriction of some $g \in C^{\infty}_{p}(M)$, then we're done. Hmmm...this is not quite true but based on the definition of germs (which says only smaller neighbourhood) and the following Proposition, albeit from Section 13, then I think we still have what we want: For $f \in C^{\infty}(S)$ and $p \in S$, there exists $g \in C^{\infty}(M)$ and $W$ open in both $S$ and $M$ s.t. $g|_W = f|_W$. QED injective.

Remark: Anyway, I use something from Section 13, so I think it's wrong of Tu to say $\iota$ is immersion and submersion, without saying that stuff from Section 13 is used...unless there's a way to do this without Section 13...

I mean, just follow the proof in Section 11...w/c technically uses techniques up to Section 8...OH WAIT NO. It uses Section 9 concepts...Eh, there's probably some up-to-Section 8 version for the Section 11 proof given that we have $k=m$.


Say that $M$ is an $n$-dimensional smooth manifold, that $S \subset M$ is an open subset, that $i: S \to M$ is the inclusion and choose $p \in S$. Since $S$ is an open subset of $M$, $S$ is also an $n$-dimensional smooth manifold (this is true because the topology on $M$ is induced from the coordinate charts on $M$, which have target $\mathbb{R}^n$). Choose a coordinate system $\phi: U \to \mathbb R^n$ for $M$, centered at $p$. By replacing $U$ with $U\cap S$ if necessary (and using the fact that $S$ is open) we may assume that $U \subset S$. We see that $\phi$ is therefore also a coordinate system for $S$ centered at $p$. Representing the inclusion map in these coordinates gives $$\phi \circ i\circ\phi^{-1} = Id,$$ where $Id:\mathbb R^n \to \mathbb R^n$ is the identity map. The differential of $Id$ at $0$ ($0$ since $\phi$ is centred at $p$) is $$Id_{*,0} :T_0\mathbb R^n \to T_0\mathbb R^n,$$ which is also the identity map.

In particular, this shows that each differential $i_{*,p}$ is both injective and surjective.

  • $\begingroup$ oh nice. so instead of directly showing each differential is injective and surjective, you just show that each differential is a composition of bijections and thus a bijection? $\endgroup$
    – BCLC
    Aug 12 at 15:48

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