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If we have a smooth vector bundle $\pi : E \to N$ and a diffeomorphism $f : M \to N$ is it true that $ f^* (\pi) :f^*E \to M$ is smooth-isomorphic to $\pi : E \to N$ ?

From the definition of the pullback we have a smooth map $g : f^*E \to E$ covering $f$. I think i can use the next result to conclude:

" A bundle morphism is an isomorphism iff its restriction to the fibers are bijective and its projection a diffeomorphism"

What i want to prove follows from this result, since $f$ is the projection of $g$ and the fiber over $p \in M$ is $\{p \} \times \pi ^{-1}(f(p)).$

Is my way of thinking correct? I ask for revision since i did not find this result explicitly in any book or exercise.

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  • $\begingroup$ You have a typo in the first sentence. $f^*E$ is a bundle over $M$. $\endgroup$ Apr 8 '18 at 1:32
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Yes, it is isomorphic, $f^*E=\{(z,y)\in E\times M$ such that $\pi(z)=f(z)\}$ define $g:f^*E\rightarrow E$ by $g(z,y)=z$ it inverse is $h(z)=(z,f^{-1}(\pi(z)))$.

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Yes. In fact this is true for homotopic maps. see Hatcher theorem 1.6

Indeed, suppose that $A,B$ are homotopy equivalent. Then there exist maps $f:A \to B$ and $g:B \to A$, so that $fg \cong id$, and $gf \cong id$, so by functoriality $f^*:Vect(B)\to Vect(A)$ is a bijection, where $f^*$ is the pullback along $f$.

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  • $\begingroup$ I know that this answer is definitely overkill here, but it's worth alerting OP to a slightly more general statement $\endgroup$ Apr 8 '18 at 1:41
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It is true.

If you are not sure about it, maybe it helps to draw a commutative diagram that follows closely the definition of vector bundle isomorphism, which is supposed to show that there does exist an inverse bundle homomorphism (with the lower map $f^{-1}$ and the upper map the obvious construction following the bijectivity)

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