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I've just read a proof that

If $M$, $N$ are smooth manifolds with boundary and $f: \partial M\rightarrow \partial N$ is a diffeomorphism then $M \cup_f N$ has a smooth manifold structure such that the obvious maps $M \rightarrow M \cup_f N$ and $N \rightarrow M \cup_f N$ are smooth imbeddings.

The proof I read uses collar neighborhoods of the two boundaries to identify a neighborhood of the common boundary in the new manifold with a product of the common boundary and an interval.

This left me wondering about the uniqueness of the smooth structure. At first I thought it must be unique and I tried to show that the identity map is smooth but I couldn't show smoothness at points on the common boundary. Then the thought of a decomposition of an exotic sphere into hemispheres made me think perhaps uniqueness isn't guaranteed. But then I wasn't sure whether the hemispheres were still $smooth$ submanifolds when you change to the exotic smooth structure. Can anyone help me out by telling me whether we always have uniqueness and if so is it easy to see that the identity map will be smooth at points in the common boundary of M and N? Thanks very much for your time.

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If you don't mind my asking, where did you read it? My gut instinct is to pull out my copy of Kosinski's Differential Manifolds and look for the result in there. Unfortunately, my copy is still at school, so I won't be able to look until tomorrow. – Jason DeVito Dec 20 '12 at 4:17
@JasonDeVito: Hi Jason. I'm the OP (I used a temporary account because I couldn't remember my password but then I was signed out for some reason). I read the proof in the new edition of Lee's "Introduction to Smooth Manifolds." – Tim kinsella Dec 20 '12 at 4:47
Dear Braudel, quite off-topic: did you choose your nickname in honour of the historian Charles Braudel? – Georges Elencwajg Dec 20 '12 at 10:57
@GeorgesElencwajg : Actually the historian Fernand Braudel! Maybe he also went by Charles? – Tim kinsella Dec 20 '12 at 17:11
Dear @Tim: no,no, my memory failed me, Braudel's first name was indeed Fernand. I find it pleasantly unexpected that your nickname refers to him : when I asked you if it did, I was pretty convinced that your answer would be "no". I am glad I was wrong... And by the way: welcome to our site! – Georges Elencwajg Dec 20 '12 at 19:00

I will try to answer your question in the case where $M,N,\partial M,\partial N$ are assumed to be compact. This additional assumption can be removed.

A smooth manifold triad is a triple $(W;V_0,V_1)$ consisting of a compact smooth $n$-dimensional manifold $W$ with boundary and two smooth compact $(n-1)$-dimensional manifolds $V_0$, $V_1$ without boundary such that $\partial W= V_0\cup V_1$ and $V_0\cap V_1=\emptyset$.

Now suppose that $(W;V_0,V_1)$, $(W';V'_1,V'_2)$ are two smooth manifold triads and $h:V_1\to V'_1$ is a diffeomorphism. The basic statement, which is made precise in the following theorem, is that one can form a well defined smooth manifold triad $(W\cup_h W';V_0,V_2')$.

Theorem: In the above situation there exists a smooth manifold structure $\mathcal{S}$ on $W\cup_h W'$ such that both inclusion maps $W\hookrightarrow W\cup_h W'$, $W'\hookrightarrow W\cup_h W'$ are diffeomorphisms onto their images. Moreover the structure $\mathcal{S}$ is unique up to a diffeomorphism leaving $V_0$, $h(V_1)=V_1'$ and $V_2'$ fixed.

Your question is a special case of this situation namely $M=W$, $N=W'$, $\partial M=V_1$, $\partial N=V_1'$ and $V_0=V_2'=\emptyset$. So the smooth structure is unique up to a diffeomorphism that leaves $f(\partial M)=\partial N$ fixed.

The theorem I quoted is theorem 1.4 from Milnor's "Lectures on the $h$-cobordism theorem".

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Thanks Dave. I looked up the proposition in Milnor's book. The proof references a theorem from Munkres' book "Elementary Differential Topology" which thankfully is also in my university's library. It seems to me that everything goes through without compactness. I think it's just easier to construct collars and bicollars when everything is compact? I'll try to piece everything together and post a proof when I have have a free moment. Thanks again. – Tim kinsella Dec 20 '12 at 17:08
If you have seen a construction of collar neighborhoods when the boundary is non-compact I would be interested in a reference. By the way: where exactly is the proof of your original statement in Lee's book? Have fun piecing everything together. :) – Dave Dec 20 '12 at 22:49
Theorem 9.25 in Lee's book (remember this is the second edition; I don't think any of this is in the first) is something he calls the "Collar neighborhood theorem." It says the boundary of any smooth manifold has a collar I.e. an open neighborhood diffeomorphic to $\partial M \times [0,1)$ (with $\partial M$ mapping to $\partial M \times 0$). What you do is choose an inward pointing vector field, flow inwards a distance depending smoothly on the starting place on the boundary, and then reparamatrize so you get the interval. The theorem I mention in the question is theorem 9.29. – Tim kinsella Dec 20 '12 at 23:30
In fact there's a remark at the bottom of page 24 in Milnors book that says compactness isn't required. – Tim kinsella Dec 20 '12 at 23:52
You are absolutely right, thanks. – Dave Dec 21 '12 at 7:37

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