Version of pseudo-isotopy $\neq$ isotopy for $(n+1)$-framings

Question

Let $M$ be a closed $n$-manifold and $\varphi$ be a self-diffeomorphisms of $M$. There is a bordism from $M$ to itself given by $M\times [0,1]$ with the identification $M \cong M \times \{0\}$ induced by $\varphi$ and the identification $M \cong M \times \{ 1\}$ induced by $\mathrm{id}_M$. If $\varphi$ is pseudo-isotopic to the identity, then this bordism is isomorphic to the identity bordism, even though $\varphi$ might not be isotopic to the identity.

I would like to know if this situation changes if we consider a framing on $TM \oplus \underline{\mathbf{R}}$ and use it to fix a framing $T(M \times [0,1]) \cong TM \oplus \underline{\mathbf{R}}$. Let $\varphi$ be a framed diffeomorphism, that is, $\varphi$ preserves the framing on $TM \oplus \underline{\mathbf{R}}$ up to a specified homotopy of framings.

Can the resulting framed bordism be isomorphic to the identity bordism without $\varphi$ being isotopic to the identity as a framed diffeomorphism?

Answer 1

Take any pseudoisotopy $\varphi\colon M\times I\rightarrow M\times I$ from the identity to a diffeomorphism $\phi$ that is not isotopic to the identity (as you mentioned, these exist). By obstruction theory, $\varphi$ can be made framed with respect to any framing of $M\times I$ (because the identity preserves any framing and $M\subset M\times I$ is an equivalence). Then you get a framing on $\phi$ in your sense by restriction. $\phi$ is not isotopic to the identity as a plain diffeomorphism, so in particular not as a framed one.