Approximation of fiber bundle isomorphisms

Question

Suppose that $p_1:E_1\to B$, $p_2:E_2 \to B$ are two $C^\infty$ fiber bundles which are $C^1$ isomorphic. That is, there exists a $C^1$ diffeomorphism $f:E_1\to E_2$ satisfying $p_2 \circ f = p_1$.

Question: Does it follow that $p_1:E_1\to B$ and $p_2:E_2 \to B$ are $C^\infty$ isomorphic?

Motivation: If $M, N$ are two $C^\infty$ manifolds which are $C^1$ diffeomorphic, then they are $C^\infty$ diffeomorphic (see, e.g., Differential Topology by Hirsch, Chapter 2). However, I am not sure if this result can be extended in the context of fiber bundles to ensure that the approximating $C^\infty$ diffeomorphisms preserve fibers.

Answer 1

This is true. Here is a high-powered proof. There are proofs in the flavor of Hirsch; I don't know a reference, but with some gusto you could prove them.

$C^k$ bundles with fiber $M$ are the same as fiber bundles with structure group $\text{Diff}^k(M)$. They are thus classified by maps $X \to B\text{Diff}^k(M)$, as long as $X$ is paracompact. Then your question follows by proving that the map $B\text{Diff}^\infty(M) \to B\text{Diff}^1(M)$ is a homotopy equivalence; because taking the loop space of this we recover the inclusion homomorphism, it suffices to show that $\text{Diff}^\infty(M) \to \text{Diff}^1(M)$ is a homotopy equivalence. But this follows from what's written in Hirsch:

Let $f: S^k \to \text{Diff}^1(M)$ be a sphere's worth of $C^1$ diffeomorphisms of $M$. By extending the smooth approximation theory in Hirsch to the case when the codomain is a Banach manifold, we may homotope $f$ to be $C^1$; this means that the induced map $f: S^k \times M \to M$ is $C^1$, not just $C^1$ in the $M$-direction. Now this is homotopic to a smooth map $f': S^k \times M \to M$; because $f(x,-)$ was a diffeomorphism and $f'$ is chosen sufficiently close to $f$, $f'(x,-)$ is also a diffeomorphism. (Diffeomorphisms are open in the $C^1$ topology.) Thus the map $\text{Diff}^\infty(M) \to \text{Diff}^1(M)$ is surjective on $\pi_k$. A similar argument in the case with boundary shows that it is injective on homotopy groups. Because $\text{Diff}^k(M)$ are metrizable manifolds, a theorem of Palais says that a weak homotopy equivalence between them is actually a homotopy equivalence, as desired.

Answer 2

As suggested by Mike Miller's comment, here is a detailed alternative proof, mainly for my own future reference. I'll prove something more general than the original question -- thus my notation will differ slightly.

All approximations will be in the strong Whitney topologies.

Lemma 1: Let $p:E\to B$ be a $C^\infty$ submersion and $s:B\to E$ a $C^1$ section ($p\circ s = \text{id}_B$). Then there exists a $C^\infty$ section $\tilde s:B \to E$ which $C^1$-approximates $s$.

Proof: We have $p\circ s = \text{id}_B\in \text{Diff}^\infty(B)$. Precomposition $C^\infty(B,E) \to C^\infty(B,B), g \mapsto p \circ g$ and inversion $\text{Diff}^\infty(B)\circlearrowleft, h\mapsto h^{-1}$ are continuous with respect to the strong $C^\infty$ topologies, and $\text{Diff}^\infty(B)\subset C^\infty(B,B)$ is open in the same topology.

Thus for all $C^\infty$ maps $g:B\to E$ that $C^1$-approximate $s$ sufficiently closely, $p \circ g \in \text{Diff}^\infty(B)$. Letting $g$ be such a map, define $\tilde s:B \to E$ via $\tilde s:= g\circ(p\circ g)^{-1}$. $\tilde s$ is a section since $p \circ \tilde s = \text{id}_B$, and $\tilde s$ approximates $s$ arbitrarily closely because of the continuity facts in the last paragraph. $\square$

Lemma 2 Let $p:E \to M$ be a $C^\infty$ submersion and $f:B\to M$ a $C^\infty$ map. Then $f^* E:= \{(b,e)\in B \times E| f(b) = p(e)\}$ is a $C^\infty$ properly embedded submanifold of $B \times E$, and $\pi_1:f^* E \to B$ is a $C^\infty$ submersion.

Proof: Since $p$ is a submersion, $p$ and $f$ are transverse maps. Thus $f \times p: B \times E \to M\times M$ is transverse to the diagonal $\Delta:= \{(m,m)\in M \times M\}$ (see problem 6-13 of the 2nd Edition of Lee's Introduction to Smooth Manifolds), and thus $f^*E = (f\times p)^{-1}(\Delta)$ is a properly embedded submanifold of dimension $\dim B + \dim E - \dim M$.

Now let $b\in B$, $v \in T_b B$, and $\gamma:(-\epsilon,\epsilon)\to B$ any smooth curve in B with $\gamma(0) = p$ and $\dot \gamma(0) = v$. Since $p$ is a submersion, we may take a local section $\sigma:U \to E$ of $p$ with $f(b)\in U$, and define $\alpha := \sigma \circ f \circ \gamma:(-\epsilon,\epsilon)\to E$ is a smooth curve with $p(\alpha(t)) = f(\gamma(t))$ for all $t$. Thus $\gamma \times\alpha:B\to f^*E$ is well-defined, and $\pi_{1*}(\dot \gamma,\dot \alpha)(0) = \dot \gamma(0) = v$. Thus $\pi_1:f^*E \to B$ is a submersion. $\square$

Theorem: Suppose $p_1:E_1\to B_1$, $p_2:E_2 \to B_2$ are two $C^\infty$ surjective submersions. Let $F:E_1\to E_2$ be a $C^1$ fiber-preserving diffeomorphism covering the $C^\infty$ map $f:B_1\to B_2$. Then there exists a $C^\infty$ fiber-preserving diffeomorphism $\tilde F:E_1\to E_2$ also covering the map $f:B_1 \to B_2$, and $\tilde F$ $C^1$-approximates $F$.

Proof: Consider the pullback $(f\circ p_1)^*E_2$, with $\pi_i: (f\circ p_1)^*E_2 \to E_i$ projection onto the $i$-th factor. Lemma 2 shows that $(f\circ p_1)^*E_2$ is a $C^\infty$ manifold and and $\pi_1: (f\circ p_1)^*E_2 \to E_1$ is a $C^\infty$ surjective submersion.

A fiber-preserving map $E_1 \to E_2$ covering $f$ is equivalent to a section of $\pi_1$: a section $s$ yields a fiber-preserving map $\pi_2 \circ s$, and a fiber-preserving map $G$ yields a section $\text{id}_{E_1}\times G$.

Let $s:E_1\to (f\circ p_1)^*E_2$ be the $C^1$ section $s:= \text{id}_{E_1}\times F$. Using Lemma 1, let $\tilde s = \text{id}_{E_1}\times \tilde F$ be a $C^\infty$ section $C^1$-approximating $s$. Then $\tilde F = \pi_2\circ \tilde s: E_1 \to E_2$ is a $C^\infty$ map, and $\tilde F$ is a diffeomorphism if $\tilde s$ is chosen sufficiently $C^1$-closely to $s$ (see first paragraph of proof of Lemma 1). Finally, $\tilde F$ is fiber-preserving covering $f$ since $p_2 \circ \pi_2 = f \circ p_1 \circ \pi_1$, so $p_2 \circ \tilde F = p_2 \circ \pi_2 \circ \tilde s = f \circ p_1 \circ \pi_1 \circ \tilde s = f \circ p_1$. $\square$

Remark: The same proof works with $C^s, C^r$ replacing $C^1, C^\infty$ for any $1 \leq s < r \leq \infty$.