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Let $M$ and $N$ be smooth manifolds and $f: M \to N$ a smooth map. Define the pullback bundle $\pi_f^*:f^*(T^*N) \to M$ as usual by $ f^*(T^*N) = \{(x,j^1_{f(x)}g) \in M \times T^*N \}$ with projection $\pi^*_f(x,j^1_{f(x)}g)=x$ then the pullback of $f$ is a smooth morphism $$f^*: f^*(T^*N) \to T^*M$$ and moreover a vector bundle morphism over the identity defined by $f^*(x,j^1_{f(x)}g) = j^1_{x}(g \circ f)$. Now the questions are:

1.) Suppose $f$ is an embedding. Is $f^*$ a surjective submersion?

2.) Suppose $f$ is a surjective submersion. Is $f^*$ an embedding?

3.) Suppose $f$ is a diffeomorphism. Is $f^*$ a diffeomorphism?

(Regarded $f^*$ just a smooth map, forgetting the additional bundle structure)

4.) Suppose $f$ is a diffeomorphism. Is $f^*$ a vector bundle isomorphism?

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up vote 1 down vote accepted

Let me adopt slightly different notation, because otherwise the symbol $f^\ast$ will be overloaded. Fix $f: M \to N$, and let $E$ be the pullback of $T^\ast N$ to $M$. Then there is a natural map $\phi: E \to T^\ast M$ given by $\phi(x, p) = (x, (df)^\ast p)$. All of your questions really come down to: what is the relation between $df$ and $d\phi$? In coordinates we have $\phi = (\phi_1, \phi_2)$ with $\phi_1(x,p) = x$ and $\phi_2(x,p) = (df)^\ast p$. So let us compute $d\phi$ in these coordinates:

$$ d\phi = \left(\begin{array}{rr} \frac{\partial \phi_1}{\partial x} & \frac{\partial \phi_1}{\partial p} \\ \frac{\partial \phi_2}{\partial x} & \frac{\partial \phi_2}{\partial p} \end{array} \right) = \left( \begin{array}{rr} \mathbb{1} & 0 \\ \frac{\partial \phi_2}{\partial x} & (df)^\ast \end{array} \right). $$ Then basic linear algebra shows that $(df)^\ast$ injective $\implies$ $d\phi$ injective, and $(df)^\ast$ surjective $\implies d\phi$ surjective. Equivalently, $df$ surjective $\implies d\phi$ injective, and $df$ injective $\implies d\phi$ surjective. The answers to your questions (1)-(4) should be more or less obvious now, so I leave the details to you.

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What is $d\phi$, is it the tangent of $\phi$? I.e $d\phi:TE \to TT^*M$ In that case (1) - (3) is true. – Mark Neuhaus Apr 14 '12 at 17:02
Yes, $d\phi$ (or $\phi_\ast$, or $T\phi$) is the differential of $\phi$. I think (4) should be true as well, since $\phi$ will be a diffeomorphism on the total spaces and it is linear on the fibers. – Jonathan Apr 14 '12 at 18:15

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