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

  • $\begingroup$ Are you asking for the definition of a manifold with boundary? Or are you asking why this quotient is not a manifold (without boundary)? $\endgroup$ – Dylan Moreland Sep 4 '11 at 0:57
  • $\begingroup$ 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 $\endgroup$ – Daniel Sep 4 '11 at 1:07
  • $\begingroup$ My new interpretation is that you want to know how to show that $X$ is a $2$-manifold with boundary. $\endgroup$ – Dylan Moreland Sep 4 '11 at 1:49
  • $\begingroup$ Ok , That will help me )= $\endgroup$ – Daniel Sep 4 '11 at 1:54
  • 3
    $\begingroup$ What's the question? $\endgroup$ – Ryan Budney Sep 4 '11 at 1:57

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.