Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.