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Let follow this definition of manifold. An n-manifold is a Hausdorff Topological space, Such That Each point you have an open neighborhood homeomorphic to the open disc $$ U^n = \left\{ {x \in R^n :\left| x \right| < 1} \right\} $$

Let this set:

$$ X = \left\{ {\left( {x,y} \right) \in R^2 /\,\,x \in \left[ { - 10,10} \right],\,y \in \left[ { - 1,1} \right]} \right\} $$

Define the quotient (10, y) related to (-10,-y) for -1

In the book ( Massey) comes out that if we consider the edges $ y=1 ,y=-1 $ Would not be a manifold under our definition. I do not understand why anyway would say that a manifold with boundary, if it refers not to ask more things or completely different. I'm just learning this quotient topology and I can not imagine the folds and stuff) =. If someone can give me advice on what to do with that, I really appreciate it


Edit: (TB) In order to give some context, let me quote the relevant passage from Massey, (assuming A basic course in algebraic topology, Springer GTM 127 was meant).

From the bottom of page 3:

The simplest example of a $2$-dimensional manifold exhibiting this phenomenon [non-orientability] is the well-known Möbius strip. As the reader probably knows, we construct a model of a Möbius strip by taking a long, narrow rectangular strip of paper and gluing the ends together with a half twist (see Figure 1.1). Mathematically, a Möbius strip is a topological space that is described as follows. Let $X$ denote the following rectangle in the plane: $$X = \{(x,y) \in \mathbf{R}^2 : -10 \leqq x \leqq +10, \;-1 \lt y \lt +1\}.$$ We then form a quotient space of $X$ by identifying the points $(10,y)$ and $(-10,-y)$ for $-1 \lt y \lt +1$. Note that the two boundaries of the rectangle corresponding to $y=+1$ and $y=-1$ were omitted. This omission is crucial; otherwise the result would not be a manifold (it would be a “manifold with boundary,” a concept we will take up in Chapter XIV [more precisely, XIV.§7, p.375ff]). Alternatively, we could specify a certain subset of $\mathbf{R}^3$ which is homeomorphic to the quotient space just described.

Unfortunately, Google managed to garble Figure 1.1, so here it is in full:

Massey, Figure 1.1 Möbius strip

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Are you asking for the definition of a manifold with boundary? Or are you asking why this quotient is not a manifold (without boundary)? –  Dylan Moreland Sep 4 '11 at 0:57
    
The difference is that the manifold with boundary it´s locally homeomorphic to $$ R^n $$ or to $$ R^n _ + $$ but how can i see geometrically this two folds )=? sorry for this stupids questions T_T –  Daniel Sep 4 '11 at 1:07
    
My new interpretation is that you want to know how to show that $X$ is a $2$-manifold with boundary. –  Dylan Moreland Sep 4 '11 at 1:49
    
Ok , That will help me )= –  Daniel Sep 4 '11 at 1:54
3  
What's the question? –  Ryan Budney Sep 4 '11 at 1:57

1 Answer 1

up vote 1 down vote accepted

So to show this is a 2 manifold with boundary you have to show that around each point there is a neighborhood that is either homeomorphic to $D^2$ or $D^2_+= \{(x,y)\in \mathbb{R} | \,\,\,\, y\geq 0, \,\,\,\, |(x,y)|<1 \}$.

Let $X$ be the described set $X / \sim$ the quotient and $\pi$ the quotient homomorphisim.

For $x \in \pi( \text{int} \, ( X )) = \text{int} \, (X)$ we are done, this set is homeomorphic to the disk. On $\text{int} \,(X)$, $\pi$ is a homeomorphism.

For $x \in \pi( (-10, 10) \times \{1\})$ consider $\pi((-10, 10) \times [1,-1))$. Similarly for the other side.

For $x \in \pi( \{10\} \times (-1,1) )$ it is more difficult. Here we have to somehow work with the twist. Let $f: [-10,-9) \cup (9,10] \times (-1,1) / \sim \,\, \to (-1,1)^2 $ be: $$ f(x,y) = \left\{ \begin{array}{lr} (x-10,y) & : x \in (9,10] \\ (x+10,-y) & : x \in [-10,-9) \end{array} \right. $$ I claim that this is continuous and bijective. Pulling $f$ back to $X$, i.e. considering $f \circ \pi : X \to (-1,1)^2$, it is continuous (this is the universal property of quotients). And it is bijective as $f \circ \pi$ is 1 to 1 except for the points that are identified where is is 2 to 1. But those points are identified so $f$ is 1 to 1 and onto. $f$ is also an open map, any open set in $X / \sim$ is the union of the images of an open sets from $X$ and $f \circ \pi$ is clearly an open map.

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