# Show that the pullback of the normal bundle of a submanifold is isomorphic to the normal bundle of the inverse image when f is transversal

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?

• For a moment, forget about the vector bundles. If $x\in f^{-1}(A)=B$, what are $T_xB$ and $(\nu_B)_x$? Jan 28, 2023 at 17:17
• $(\nu_B)_x=\frac{T_xX}{T_xB}$, and because $codim(A)=codim(B)$ there is a vector space isomorphism with $(f^*(\nu_A))_x$. Am I right? Jan 28, 2023 at 20:37
• 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. Jan 28, 2023 at 20:39
• 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. Jan 28, 2023 at 22:06

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 $$f^*(N_A X)$$.