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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.

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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.

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  • $\begingroup$ 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. $\endgroup$ – Max Lipton Jul 1 '17 at 20:18

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