Consider smooth fiber bundles $P_i:E_i\to B_i,\quad i=1,2\quad$ with fiber $F$. Let $\tilde f:E_1\to E_2$ be a smooth bundle map. That is a smooth map which preserves the fiber $F$. Let $f:B_1\to B_2$ be the unique map such that $P_2\circ\tilde f=f\circ P_1$.

If $f$ is a diffeomorphism, we can see that $\tilde f$ is a bijection. How can we see that $\tilde f^{-1}$ is smooth?

Moreover, if $\tilde f$ is a diffeomorphism we can see that $f$ is a bijection. How can we see that $f^{-1}$ is smooth?

Observation: If $g:X\to Y$ is a continuous bijection between a compact space $X$ and a Hausdorff space $Y$, then $f^{-1}$ is continuous. However, neither $E_1$, nor $B_1$ are compact.

Addendum: I am reading Morita's "Geometry of Differential Forms." His definitions are like the following.

Def $1$: (Differentiable fiber bundle) Let $F$ be a smooth manifold. Suppose $E$ and $B$ are smooth manifolds and $P:E\to B$ is a smooth map. We say that $K=(E,P,B,F)$ is a differentiable fiber bundle with fiber $F$ is it satisfies the local triviality condition. That is $(\forall b\in B) (\exists b\in U_{\text{open}}\subset B)(\exists \phi:P^{-1}(U)\tilde = U\times F)(\forall u\in P^{-1}(U))(P(u)=P_1\circ\phi(u))$, where $P_1$ is the projection onto the first factor.

Def $2$: (Bundle map) Let $P_i:E_i\to B_i,\quad i=1,2\quad$ be differentiable fiber bundles with the same fiber. By a bundle map from $P_1$ to $P_2$ we mean smooth maps $\tilde f:E_1\to E_2$ and $f:B_1\to B_2$ such that $P_2\circ\tilde f=f\circ P_1$ and such that the restriction of $\tilde f$ to any fiber $P_1^{-1}(b)$ is a diffeomorphism.

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    $\begingroup$ Take each bundle to be the trivial real line bundle over the real line, with projection to the first factor. The mapping $\tilde{f}(x, v) = (x, v^{3})$ is a smooth map that preserves fibres and covers the identity map (a diffeomorphism),but is not itself a diffeomorphism. That is, there's presumably some "fine print" in the definitions of "bundle map" and/or "fibre-preserving" that you're not using. If you go back to the complete definitions, does it help? If not, could you please add the missing conditions to your question? Thanks. :) $\endgroup$ May 7, 2015 at 12:04
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    $\begingroup$ I beg your pardon, @user86418, $\tilde f$ restricted to a fiber is not a diffeomorphism. $\endgroup$ May 7, 2015 at 13:32
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    $\begingroup$ This definition of bundle map is quite restrictive . For example the zero morphism between two vector bundles on the same base would not be a bundle map with this definition. $\endgroup$ May 7, 2015 at 13:55
  • $\begingroup$ @Student: The idea is that a bundle map $\tilde{f}$ is (by definition) a diffeomorphism in the fibre directions (the hypothesis omitted in the initial wording), and so is a diffeomorphism if and only if the induced map of base spaces is a diffeomorphism. Since you already have bijectivity, the main issue is, as you note, smoothness of the inverse maps. Smoothness of $\tilde{f}^{-1}$ clearly implies smoothness of $f^{-1}$. Conversely, since smoothness is local, it's sufficient to restrict your attention to a trivializing neighborhood (where the bundle is a product). Does that help? $\endgroup$ May 7, 2015 at 14:09
  • $\begingroup$ Your suggestion helps me in part, @user86418. I see that smoothness of $\tilde f^{-1}$ implies smoothness of $f^{-1}$. Conversly, consider a point $b\in B_2$, a trivialization $x_2: P_2^{-1}(V)\to V\times F$ around $b$, and a trivialization $x_1:P_1^{-1}(U)\to U\times F$ around $f^{-1}(b)$. This gives us $P_1\circ x_1^{-1}\circ\tilde f^{-1}=f^{-1}\circ P_2\circ x_2^{-1}$. All the maps involved in the last identity are smooth except possibly $\tilde f^{-1}$. How can we conclude? $\endgroup$ May 7, 2015 at 16:28

1 Answer 1


In a trivializing neighborhood $U \times F$ (using the notation of your comment), we may write $$ \tilde{f}(b, e) = \bigl(f(b), \phi(b, e)\bigr) $$ for some smooth mapping $\phi:U \times F \to F$ that is a diffeomorphism on each fibre $\{b\} \times F$. The Jacobian has a block decomposition $$ D\tilde{f}(b, e) = \left[\begin{array}{@{}cc@{}} Df(b) & D_{1}\phi(b, e) \\ 0 & D_{2}\phi(b, e) \\ \end{array}\right], $$ in which $D_{2}\phi$ is invertible at each point $u = (b, e)$ by the definition of a bundle map. It follows that $D\tilde{f}$ is invertible at each point if and only if $Df(b)$ is invertible at each point.

You may be able to express this in matrix-free form, but matrices do highlight the main idea: $\tilde{f}^{-1}$ "is a diffeomorphism in the fibre direction" by hypothesis, so any failure of smoothness must occur "in the base direction". This happens if and only if $f^{-1}$ is not smooth, since $P_{1} \circ x_{1}^{-1} \circ \tilde{f}^{-1} = f^{-1} \circ P_{2} \circ x_{2}^{-1}$.


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