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I have a question about a proof in John Lee's Introduction to Topological Manifolds.

Suppose $M$ and $N$ are two topological $n$-manifolds with nonempty boundary (for reference, the definition I am using is: $M$ and $N$ are second countable Hausdorff spaces such that each point has an open neighborhood homeomorphic to either an open set in $\mathbf{R}^n$ or an open set in $\mathbf{H}^n = \{ (x_1,\ldots,x_n) \in \mathbf{R}^n: x_n \geq 0\}$). If $\partial M$ and $\partial N$ denote the manifold boundaries of $M$ and $N$, suppose $h: \partial N \to \partial M$ is a homeomorphism. We may attach $M$ and $N$ along their boundaries by forming the adjuction space

$$ M \cup_h N = \frac{M \sqcup N}{\sim}$$

where $\sim$ is the relation generated by identifying $x\in \partial N$ with $h(x)$. The theorem is:

With $M$, $N$, and $h$ as above, $M \cup_h N$ is an $n$-manifold (without boundary).

The proof first proceeds to show that each point of the adjuction space has a euclidean neighborhood. If $q: M \sqcup N \to M \cup_h N$ denotes the quotient map, then Lee puts $S=q(\operatorname{Int} M \cup \operatorname{Int} N)$ (where $\operatorname{Int}$ denotes the manifold interior) and shows that each point of $M \cup_h N -S$ has a euclidean neighborhood. I am OK with this argument.

Now, I quote:

Suppose $s\in S$, and let $y \in \partial N$ and $x=h(y) \in \partial M$ be two points in the fiber $q^{-1}(s)$. We can choose coordinate charts $(U,\phi)$ for $M$ and $(V, \psi)$ for $N$ such that $x\in U$ and $y\in V$, and let $\hat{U}=\phi(U)$, $\hat{V} = \psi(V) \subseteq \mathbf{H}^n$. It is useful in this proof to identify $\mathbf{H}^n$ with $\mathbf{R}^{n-1}\times [0,\infty)$ and $\mathbf{R}^n$ with $\mathbf{R}^{n-1} \times \mathbf{R}$. By shrinking $U$ and $V$ if necessary, we may assume $h(V \cap \partial N) = U \cap \partial M$, and that $\hat{U} = U_0 \times [0,\epsilon)$ and $\hat{V} = V_0 \times [0,\epsilon)$ for some $\epsilon>0$ and some open subsets $U_0,V_0 \subseteq \mathbf{R}^{n-1}$.

My question: How does one choose the coordinate domains $U$ and $V$ to satisfy these two conditions simultaneously? I know how to choose $U$ and $V$ such that $h(V \cap \partial N) = U \cap \partial M$, and I know how to choose $U$ and $V$ such that their images under the coordinate maps are equal to sets of the form $U_0 \times [0,\epsilon)$ and $V_0 \times [0,\epsilon)$, but I don't know how to pick $U$ and $V$ such that both of these conditions are satisfied at the same time.

EDIT: Let me be a bit more specific. We are to choose the coordinate domains (by shrinking if necessary) to satsify the following two conditions simultaneously:

(a) $h(V \cap \partial N) = U \cap \partial M$,

(b) $\hat{U} = U_0 \times [0,\epsilon)$ and $\hat{V} = V_0 \times [0,\epsilon)$.

Now, assuming that I have adjusted $U$ and $V$ to satisfy (a), then I may adjust them further to satisfy (b). But then how do I know that $U$ and $V$ still satisfy (a)? I can be more specific by what I mean by "adjust" if neeeded.

Am I misreading the proof, or am I missing something glaringly obvious?

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2 Answers 2

up vote 2 down vote accepted

Here's a way to see it. Choose charts $(U,\varphi)$ and $(V,\psi)$ such that $x\in U$ and $y\in V$, and let $U_1 = U\cap \partial M$ (which is a neighborhood of $x$ in $\partial M$) and $V_1 = V\cap \partial N$ (a neighborhood of $y$ in $\partial N$). After shrinking $U$ if necessary, we may assume that $h(U_1)\subset V_1$. (Because $U_1\cap h^{-1}(V_1)$ is a neighborhood of $x$ in the subspace topology $\partial M$ inherits from $M$, there is an open subset $\widetilde U\subset M$ such that $U_1\cap h^{-1}(V_1) = \widetilde U\cap\partial M$. Replacing $U$ by $U\cap \widetilde U$ does the trick.) Let $U_0\subset U_1$ be a neighborhood of $x$ that's precompact in $U_1$, and let $V_0 = h(U_0)\subset V_1$, a precompact neighborhood of $y$ in $V_1$.

Now let $\widehat U=\varphi(U)$, $\widehat V=\psi(V)$, considered as subsets of $\mathbb H^n = \mathbb R^{n-1}\times [0,\infty)$. Since $\varphi(\overline {U_0})$ is a compact subset of $\mathbb R^{n-1}\times \{0\}$, there is some $\varepsilon_1>0$ such that $\varphi(\overline {U_0})\times[0,\varepsilon_1)\subset \widehat U$. Similarly, there's $\varepsilon_2>0$ such that $\psi(\overline {V_0})\times[0,\varepsilon_2)\subset \widehat V$. Now just take $\varepsilon = \min(\varepsilon_1,\varepsilon_2)$, and replace $U$ by $\varphi^{-1}\big(\varphi(U_0)\times[0,\varepsilon)\big)$ and $V$ by $\psi^{-1}\big(\psi(V_0)\times[0,\varepsilon)\big)$.

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Wow, thank you for the response Dr. Lee. One more quick question and then I've got it: Why does compactness of $\phi( \overline{U}_0)$ in $\mathbf{R}^{n-1} \times \{ 0\}$ imply the existence of $\epsilon_1$? –  John Myers May 18 '13 at 22:24
    
Oh, scratch that last question. I see why! –  John Myers May 18 '13 at 22:33
    
Since the Euclidean topology on $\mathbb R^{n-1}\times[0,\infty)$ is the same as the product topology, for each $p\in\varphi(\overline{U_0})$ there exist a neighborhood $U_p$ of $p$ in $U_1$ and an $\varepsilon_p>0$ such that $U_p\times [0,\varepsilon_p)\subset \widehat U$. By compactness, finitely many sets $U_{p_1},\dots,U_{p_k}$ cover $\varphi(\overline{U_0})$, and we can take $\varepsilon_1 = \min(\varepsilon_{p_1},\dots,\varepsilon_{p_k})$. –  Jack Lee May 18 '13 at 22:44
    
Hi, Jack :) Don't you just hate having to come up with solutions of your own problems? –  Ted Shifrin May 18 '13 at 22:53
    
Hi Ted -- I mostly ignore it when people ask for solutions to the problems, because I think they'll learn more by grappling with them on their own, or with help from other learners. But when they find my explanations in the book unclear, I do feel a bit of an obligation to try to help... –  Jack Lee May 19 '13 at 4:37

I take it you're worried about having the $\epsilon$'s be the same? You can rescale by a linear map in the $x_n$ coordinate.

OK, if the first condition no longer holds, shrink either $U_0$ or $V_0$ to make it hold.

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No, unfortunately that's not my issue. I will edit the question so that it's more clear. –  John Myers May 17 '13 at 3:38
    
Can you elaborate more on your second comment? I'm not immediately seeing how I can make $U$ and $V$ satisfy the first condition by shrinking $U_0$ and $V_0$. –  John Myers May 17 '13 at 4:07
    
Remember that $U_0\times\{0\}=\phi(U\cap\partial M)$, etc. So we're just going to rig it so that $\phi\circ h\circ \psi^{-1}$ carries $V_0$ to $U_0$. –  Ted Shifrin May 17 '13 at 4:11
    
I apologize for prodding for more details, but I still am not seeing how to choose the correct $U$ and $V$ based off of shrinking $U_0$ and $V_0$. I understand the rest of the proof, it's just this small technicality that is giving me trouble. It's an intuitive proof; essentially what we are trying to do is align the boundaries of $M$ and $N$ so that we can define a map $U \sqcup V \to \mathbf{R}^n$ that passes to the quotient (giving us our coordinate map of $s$). Continued... –  John Myers May 17 '13 at 16:02
    
Assuming $U$ and $V$ have been selected to satisfy the two requirements, we write $\phi=(\phi_0,\phi_1)$ and $\psi=(\psi_0,\psi_1)$ and note that the first two coordinate functions $\phi_0$ and $\psi_0$ provide homeomorphisms $U \cap \partial M \to U_0$ and $V \cap \partial N \to V_0$, resepectively. Then $(\phi_0 | U \cap \partial M) \circ h \circ (\psi_0| V \cap \partial N)^{-1}$ is the map you mentioned in your last comment mapping $V_0$ homeomorphically onto $U_0$. –  John Myers May 17 '13 at 16:06

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