TITLE: If two smooth manifolds are homeomorphic, then their stable tangent bundles are vector bundle isomorphic
QUESTION [9 upvotes]: I am currently reading Kervaire-Milnor's paper "Groups of Homotopy Spheres I", Annals of Mathematics, and I am trying to prove (or disprove) the following result. The more elementary the proof, the better.

If two smooth manifolds are homeomorphic, then their stable tangent
bundles (i.e. the Whitney sum of the tangent bundle with the trivial
line bundle) are vector bundle isomorphic.

I am trying to prove this as an intermediate step to give an alternative proof for KM's Theorem 3.1: Every homotopy sphere is $s$-parallelizable.

REPLY [8 votes]: Let me add something to Michael Albanese's great answer to see this question in a broader context.
Novikov proved that the rational Pontryagin classes are homeomorphism invariants (in fact, this was one of the achievements for which he received the Fields medal in 1970). The integral Pontryagin classes, however, are not invariant under homeomorphism, see Chapter 4.4 of "The Novikov Conjecture" by Kreck and Lück.
Some polynomials in the Pontryagin classes are even homotopy invariant: for instance, $p_1$ of a closed oriented $4$-manifold $M$ agrees (by Hirzebruch's signature theorem) with $3\sigma(M)$ times the fundamental class in cohomology (where $\sigma(M)$ denotes the signature of $M$), which is invariant under homotopy equivalence.
The famous Novikov conjecture asks whether certain so-called "higher signatures" are also invariant under homotopy equivalence. It is one of the most important open questions in topology.