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Let $f:M' \to M$ be a map between smooth manifolds. Let $S \subset M$ be a submanifold, and let $T$ be a tubular neighborhood of $S$, ie. $T$ is diffeomorphic to the normal bundle of $S$ in $M$. If $f$ is transversal to $S$ and $T$, $f^{-1}(T)$ is a tubular neighborhood of $f^{-1}(S)$ provided $T$ is small enough.

This fact was used in P.69 of Bott, Tu, but I cannot come up with a proof of this. Ideally I would like to construct a diffeomorphism between $f^{-1}(T)$ and the normal bundle of $f^{-1}(S)$ directly, but I have not succeeded. I noticed the authors explicitly stated $T$ needs to be sufficiently small, this lead me to think maybe we have to approach this locally, perhaps allowing us to consider everything to be in Euclidean space, but I have no clear idea how this is done. Any help is appreciated!

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Perhaps I'm missing your point, but the key issue is that when $f$ is transverse to $S$, not only is $f^{-1}(S)$ a submanifold of $M'$, but also one has $$N_{f^{-1}(S)/M'} = f^*N_{S/M}.$$ (Then it's just a matter of applying the tubular neighborhood theorem.) Indeed, the bundle map $df$ maps $N_{f^{-1}(S)/M'}$ isomorphically to $N_{S/M}$. (Although this is often not stated explicitly, $f\pitchfork S \iff df_x$ maps $T_xM'$ surjectively to $T_{f(x)}M\big/T_{f(x)}S \cong \big(N_{S/M}\big)_{f(x)}$ for each $x\in f^{-1}(S)$.)

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  • $\begingroup$ Thank you for your response, but then how can I show $f^{-1}(T)$ (the preimage, not the pullback bundle) is diffeomorphic to $f^*N_{S/M}$? $\endgroup$
    – Chi Cheuk Tsang
    Commented Jul 16, 2016 at 0:54
  • $\begingroup$ I don't understand what you're asking. You use the tubular neighborhood theorem along with the bundle isomorphism. $\endgroup$
    – Ted Shifrin
    Commented Jul 16, 2016 at 1:06
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    $\begingroup$ Look carefully at the definition of tubular neighborhood on p. 65 of Bott/Tu. $\endgroup$
    – Ted Shifrin
    Commented Jul 16, 2016 at 16:25
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    $\begingroup$ You needn't have a precise $\epsilon$-neighborhood (in some Riemannian metric) in order to have a tubular neighborhood. You do want normal bundles, but the point of my answer is that the normal bundle of $f^{-1}(S)$ is the pullback of the normal bundle of $S$, and that's all it takes. Take a small neighborhood of the zero-section and transplant it by the bundle isomorphism. $\endgroup$
    – Ted Shifrin
    Commented Jul 17, 2016 at 0:37
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    $\begingroup$ Provided that $V$ is small enough to be diffeomorphic to a neighborhood of $f^{-1}(S)$ in $M'$, this works perfectly. Otherwise, shrink to make it so (and take $U$ to correspond to it). Chasing all the diffeomorphisms (there are three), yes, $f$ will map the neighborhood of $f^{-1}(S)$ in $M'$ diffeomorphic to $V$ to the neighborhood of $S$ in $M$ diffeomorphic in $U$ diffeomorphically (because it's a composition of three diffeomorphisms). $\endgroup$
    – Ted Shifrin
    Commented Jul 17, 2016 at 2:37

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