# Does the smooth homotopy between diffeomorphisms have to be through diffeomorphisms in order to induce the same equivalence class of cobordisms?

I'm currently reading through Kock's notes on TQFTs, found here http://mat.uab.es/~kock/TQFT/FS.pdf, and I assume all of his definitions are standard.

Suppose we have two diffeomorphic, closed, compact, smooth $(n-1)$-manifolds $\Sigma_0$ and $\Sigma_1$, with $\psi_1, \psi_2: \Sigma_0 \to \Sigma_1$ being diffeomorphisms. These diffeomorphisms induce cobordisms from $\Sigma_0$ to $\Sigma_1$ via the cylinder construction. (1.1.4 in the notes)

If $\psi_1$ and $\psi_2$ are smoothly homotopic, then we have a smooth map $F: \Sigma_0 \times I \to \Sigma_1$, where $F(x,0) = \psi_1(x)$ and $F(x,1) = \psi_2(x)$ for all $x \in \Sigma_0$. There is an obvious map $\Sigma_0 \times I \to \Sigma_1 \times I$ given by $(x,t) \mapsto (F(x,t),t)$. The notes claim that this is an equivalence of cobordisms from $\Sigma_0$ to $\Sigma_1$, after appropriate identifications of the boundary to make them actual cobordisms from $\Sigma_0$ to $\Sigma_1$.

However, I don't think $(x,t) \mapsto (F(x,t),t)$ is a diffeomorphism, unless $F_t$ is a diffeomorphism for all $t$, which I haven't assumed. Must I assume this?

Edit: The homotopy need not be through diffeomorphisms (isotopy) but the homotopy map is not a priori guaranteed to be a diffeomorphism.

## 1 Answer

The answer here https://mathoverflow.net/questions/155380/mapping-class-group-vs-automorphism-group-in-cobordism-category (really just the first stanza) implies that for any two non-homotopic automorphisms of a surface the cyllinder construction gives distinct cobordisms. The statement in 1.1.5 that all automorphisms induce equivalent cobordisms seems grossly wrong to me (EDIT: I guess I am assuming here that the precise cyllinder is one side the inclusion $\Sigma_0$ and on the other side the diffeomorphism. If the author means the map where both sides are the diffeomorphism, then these should be equivalent for a rather trivial reason. Namely your diffeomorphism is $\phi \circ \psi ^{-1}$ where $\phi$ and $\psi$ are your diffeo's and don't touch the $\times I$ bit).

In general two automorphisms induce equivalent cobordisms if they are "pseudo-isotopic". This paper of Browder http://www.ams.org/journals/tran/1967-128-01/S0002-9947-1967-0212816-0/S0002-9947-1967-0212816-0.pdf gives an example of a map which is homotopic to the identity, but not pseudo-isotopic to the identity which gives a precise counterexample to the thing you are trying to do.

With no negative implication towards the notes you are using, the notes by Freed available here http://www.ma.utexas.edu/users/dafr/M392C-2012/ are excellent.

• Thanks for the reply. This confirms the answer to my question is no in general as I suspected. However, pseudo-isotopy is a weaker condition than isotopy so the true conditions for maps inducing the same cobordism is between smooth homotopy and smooth isotopy. – Max Lipton Jul 1 '17 at 20:18