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Let $f\colon X\rightarrow Y$ be transversal to $A\subset Y$ submanifold of $Y$. I've been trying to show that the pullback of the normal bundle of $A$ $f^*(\nu_A)$ is isomorphic to the normal bundle of the inverse image $\nu_{f^{-1}(A)}$. I understand why the inverse image $f^{-1}(A)$ is a submanifold of the same codimension as $A$, and I know the condition of transversality. My idea was to define some diffeomorphism between the total spaces $E(\nu_{f^{-1}(A)})=\{(x,v)\in f^{-1}(A)\times (T_{x}(f^{-1}A))^{\perp}\}$ and $E(f^*(\nu_A))=\{(x,v)\in f^{-1}(A)\times (T_{f(x)}A)^{\perp}\}$ and then check the isomorphism in the fibers, but I'm stuck. I guess I have to use somewhere that for $w\in\frac{T_{f(x)}Y}{T_{f(x)}A}$, $w=0$ iff $w\notin f_*(T_xX)$. Can someone help?

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    $\begingroup$ For a moment, forget about the vector bundles. If $x\in f^{-1}(A)=B$, what are $T_xB$ and $(\nu_B)_x$? $\endgroup$ Jan 28, 2023 at 17:17
  • $\begingroup$ $(\nu_B)_x=\frac{T_xX}{T_xB}$, and because $codim(A)=codim(B)$ there is a vector space isomorphism with $(f^*(\nu_A))_x$. Am I right? $\endgroup$ Jan 28, 2023 at 20:37
  • $\begingroup$ But the existence of fiberwise isomorphisms doesn't imply at all the existence of a isomorphism in the category of vector bundles, for example $T(S^1)\neq M$ where $M$ is the Möbius strip. $\endgroup$ Jan 28, 2023 at 20:39
  • $\begingroup$ But you haven't answered my questions. How are $f$ and the transversality hypothesis relevant? You might start by thinking about multivariable calculus and the normal vectors to the submanifold defined by taking the preimage of a regular value. $\endgroup$ Jan 28, 2023 at 22:06

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We first construct a map $\phi: N_{f^{-1}(A)}Y\rightarrow f^*(N_A X)$, then prove it to be an isomorphism.

For every fixed point $x\in f^{-1}(A)$ and $[v]\in (N_{f^{-1}(A)}X)_x=T_x X/T_x f^{-1}(A)$ represented by $v\in T_x X$, we define $\phi([v])$ by $[f_*(v)]\in (f^*(N_A Y))_x=(N_A Y)_{f(x)}=T_{f(x)}Y/T_{f(x)} A$. Then one can easily prove that $\phi$ constructed above is well-defined and injective (notice here we are using that $v\in T_x f^{-1} A$ iff $f_*(v)\in T_{f(x)} A$, similar to what you have mentioned in the description). To show $\phi$ is surjective, it is equivalent to proving $f_*(T_x X)+T_{f(x)} A=T_{f(x)} Y$, which is just another way saying that $f$ is transversal to $A$!

Therefore, when $f$ is transversal to $A$, the map $\phi$ gives an isomorphism between $N_{f^{-1}(A)}Y$ and $f^*(N_A X)$.

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