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Here is my second question on understanding jets better:

For a smooth manifold $M$ the set of jets $J^1_0(\mathbb{R},M)$ is the same as the Tangent bundle $TM$. This implies that any equivalence class $j_0^1f$ of curves $f:\mathbb{R} \rightarrow M$ is a tangent vector at $f(0)$.

Now suppose $N \subset M$ is a submanifold of $M$. Then we can look on it from two different angles:

First $N$ is a manifold in its own right. In that case the jet class $j_0^1f \in J^1_0(\mathbb{R},N)$ contains smooth maps $f: \mathbb{R} \rightarrow N$ only.

But if we look on $N$ as a submanifold of $M$, then a jet class $j_0^1g \in J^1_0(\mathbb{R},M)$, that is tangent to $N$, defining the same tangent vector as $j_0^1f$, is not 'the same' as $j_0^1f$ as a set of curves.

Although $j_0^1g$ is tangential to $N$, it contains curves that intersect $N$ only in $g(0)$ just having the same first order derivative in $g(0)$.

So seen this way, although $j_0^1g$ and $j^0_1f$ define the same tangent vector in $f(0)$ they are not the same as sets of curves.

Is this right?

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Now to show that a jet $j_0^1g \in J^1_0(\mathbb{R},M)$ is actually tangent to a submanifold $N$, is it sufficient to show that there is a representative $h \in j^1_0g$ and an open neighbourhood $U$ of zero in $\mathbb{R}$ such that $h$ is locally a curve in $N$, that is $h : U \rightarrow N$ ?

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It may (though I don't guarantee it) be helpful to think about the class of all smooth functions $\pi:M\to N$ such that denoting the submanifold inclusion $i:N\to M$ we have $\pi\circ i = Id$. –  Willie Wong Jan 16 '12 at 11:57
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up vote 1 down vote accepted

The magic concept here is functoriality.

1) Jets and tangent spaces are essentially useless if you do not say how they react to differentiable morphisms between manifolds.
More precisely, consider a $\mathcal C^1$- map $\phi: N\to M$ between two manifolds $N, M$, sending $n$ to $m$ : $\phi (n)=m$. Then it induces induces a linear map
$$T_n(\phi):T_n(N)\to T_m(M): j^1_n(f)\mapsto j^1_m(\phi\circ f)$$
(Of course there is something to prove: namely that if $f$ and $g$ have the same jet at $n$, so have $\phi \circ f$ and $\phi \circ g$ at $m$. Also, I have slightly changed your notation to emphasize the point of the manifold where you take the jet.)

2) This applies of course in your case, when $\phi$ is the inclusion $i:N\hookrightarrow M$ of the submanifold $N$ into $M$.
Suppose you have a curve $f:\mathbb R\to N$. Its jet consists of a family $j^1_{N,n}(f)$ of many brothers and sisters $g:\mathbb R\to N$.
Now the effect of $T_n(i)$ on $j^1_{N,n}(f)$ is to send it to $j^1_{M,n}(f)$.
Is it the same family?
No!
The old brothers and sisters remain brothers and sisters but the family is enlarged:
there are new brothers and sisters for $f$, namely the curves $h:\mathbb R\to M$ which do not necessarily remain in $N$ after their visit at $n$ but nevertheless are tangent to $f$.
An easy example: $N=\mathbb R\times 0\subset M=\mathbb R^2$, $f(t)=(t,0) ,\; h(t)=(t,t^2) $ .

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ok. But what about the last part of my question? –  Mark Neuhaus Jan 16 '12 at 13:17
    
The answer to the last part of your question is "yes" because in the notation of your question we have $j^1_0(g)=T_n(i)(j^1_0(f))$. To be extremely precise, we identify $T_{N,n}$ with the subspace $T_n(i)(T_{N,n})$ of $T_{M,n}$. –  Georges Elencwajg Jan 16 '12 at 13:52
    
Ok. Thanks for your help! –  Mark Neuhaus Jan 16 '12 at 13:57
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