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M' is a manifold with boundary. One can attaching a handle $h:=D^k\times D^{n-k}$ along $f:S^{k-1}\times D^{n-k}\rightarrow M'$ forming $M=M'\cup_f h$. Suppose f is homotopic with f', and that f,f' are smooth embeddings. Is there a theorem concluding $M=M'\cup_f h$ is diffeomorphic with $M=M'\cup_f' h$ by imposing some sort of restriction on the homotopy between f and f'(Or ideally, no restriction)?

Actually I'm just asking for theorems that make attaching handles more flexible...Any theorem helpful is quite welcomed.


By the way, I wonder whether there is such kind of more general theorem: $f_0,f_1:N\rightarrow M$ are embeddings(topologically or smoothly), $F:N\times I\rightarrow M$ is a homotopy connecting $f_0, f_1$ satisfying some condition(for example $F_t$ is an embedding for any t). Suppose $N\subset N'$ and there is an embedding $g_0:N'\rightarrow M$ extending $f_0$. Then F extends to a $G:N'\times I\rightarrow M$ with $G_0=g_0$, $G_1$ still an embedding .

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It is rarely a good idea to use abbreviations in titles. –  Mariano Suárez-Alvarez Aug 3 '11 at 16:24
    
I was being lazy... Fixed. Thanks. –  Honglu Aug 3 '11 at 16:36
    
I am pretty sure the answer is, yes, homotopic attaching maps give you diffeomorphic manifolds, but I'm at home and can't look it up. Have you checked any standard references on handle theory? –  Grumpy Parsnip Aug 3 '11 at 16:45
    
Actually no. What are the standard references on handle theory? I didn't learn this kind of theory systematically. –  Honglu Aug 3 '11 at 16:56
    
lethe: I agree; I too have been groping my way thru the dark trying to learn these topics. –  gary Aug 3 '11 at 16:59

1 Answer 1

Your question can be subject to interpretation, but here's a proof that you cannot hope for too strong a result.

Take the simplest example you can think of: $M'$ is the $n$-ball, and you'll add a $n$-handle (or is it $n+1$?), that is another $n$-ball. Your embedding $f$ is a diffeomorphism of the $(n-1)$-sphere and $M = D^n \sqcup_f D^n$.

Of course, it is quite easy to show that all diffeomorphisms $S^{n-1} \to S^{n-1}$ reverting the orientation are homotopic (they are even isotopic through homeomorphisms), and for your favorite one, $M$ is just $S^n$. So your question is: are all these spaces diffeomorphic? As the diffeomorphisms are isotopic through homeomorphisms, $M$ is certainly homeomorphic to $S^n$.

Here comes the punchline: you can construct in that way manifolds that aren't diffeomorphic to the sphere (so called exotic spheres). It's a celebrated result of Milnor (1956). We have something of a classification of those beasts (Kervaire-Milnor, 1963), and we know that all manifolds homeomorphic to the spheres (of dimension $\neq 4$) are obtained by this construction (the construction is called "clutching", and this result is due to Smale. (In dimension 4, we know that no exotic sphere can be construct that way (a result of Cerf's) but we still don't know is an exotic sphere exists at all).

In all cases, that puts some serious restrictions on the result you dream about.

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Interesting! But to put the differentiability on the homotopy is okay for my purpose. –  Honglu Aug 3 '11 at 17:11
    
So you mean your homotopy is an isotopy of differentiable embeddings? –  PseudoNeo Aug 3 '11 at 17:20
    
@lethe: you can without loss of generality assume your homotopy to be differentiable by Whitney's approximation theorem, so PseudoNeo's answer would still be valid in this case. Having a differentiable isotopy of differentiable embeddings should work, however... –  Alexander Thumm Aug 3 '11 at 17:39
    
Ahh yes, only differentiability wouldn't work...Then assume isotopy of differentiable embeddings...Although it is quite restrictive, having one is better than none. –  Honglu Aug 4 '11 at 11:40

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