1
$\begingroup$

Let $S_1, S_2$ be two topological manifolds without boundary and let $D_1,D_2$ be two disks embedded in $S_1$ and $S_2$ respectively. Define $X_i:=S_i\setminus\text{int}(D_i)$. Let $Y$ be the space by gluing $\partial D_1$ to $\partial D_2$. I want to show that $Y$ is topological manifold. I have trouble showing this for points on the boundary components. Let $x\in \partial D_1$. Then $x\sim \tilde{x}\in\partial D_2$. I want to find an open neihbourhood of $x$ homeomorphic to $\mathbb{R}^2$. How can I do this?

$\endgroup$
  • $\begingroup$ Well, for one thing, you will have to look at the definition of quotient spaces. Otherwise, how will you know what any neighbourhood of $x$ looks like? So, what is your definition of quotient space topology? $\endgroup$ – Arthur Mar 19 at 18:45
  • $\begingroup$ Let $h:\partial D_1\to\partial D_2$ be a homeomorphism. Then the quotiënt space is $X_1\cup X_2/\{x\sim h(x)\}$ $\endgroup$ – user408856 Mar 19 at 18:49
  • $\begingroup$ Sure, those are the points in the quotient space. That's not what I asked about. $\endgroup$ – Arthur Mar 19 at 19:17
  • $\begingroup$ Oh I meant, endowed with the quotient topology $\endgroup$ – user408856 Mar 20 at 7:23
  • $\begingroup$ And I repeat my original question: So, what is your definition of quotient space topology? $\endgroup$ – Arthur Mar 20 at 7:26
0
$\begingroup$

I'm going to assume that you know all about quotient spaces (if you don't then you'll have to learn that because otherwise no answer to your question will make sense).

Another thing you need is the collar neighborhood theorem from differential topology, applied to the boundaries of $X_1$ and $X_2$. That theorem says that there exist neighborhoods $N_1$ of $\partial X_1$ and $N_2$ of $\partial X_2$ and diffeomorphisms $f_1 : N_1 \to \partial X_1 \times [0,1)$ and $f_2 : N_2 \to \partial X_2$.

Now choose a diffeomorphism $g : \partial X_1 \to \partial X_2$. And then we have the quotient topological space $Y$ together with the quotient map $q : X_1 \cup X_2 \to Y$ obtained by identifying each $x \in \partial X_1$ with $g(x) \in \partial X_2$, so $q(x)=q(g(x))$.

Let me alter your notation slightly: I'll choose $x_1 = \partial D_1 = \partial X_1$ and $x_2 = g(x_1) \in \partial D_2 = \partial X_2$, which corresponds to the point $x = [x_1] = [x_2] \in Y$ (here I use the notation $[\cdot]$ to denote the corresponding point in the quotient space; I'm unsure whether this is what you intend in your question when you put the $\tilde{}$ symbol over something).

So now I have to describe a manifold chart in $Y$ for the point $x$. To do this, I'll choose a manifold chart in $\partial X_1$ around $x_1$, i.e. an open subset $U_1 \subset \partial X_1$ containing $x_1$ and a diffeomorphism $\phi_1 : U_1 \to B$, where $B$ is the unit open ball in Euclidean space. From this I get a manifold chart in $\partial X_2$ around $x_2$, namely $U_2 = g(U_1)$ and $\phi_2 = \phi_1 \circ g^{-1} : X_2 \to B$.

In $Y$ define the open chart around $x$ as follows: $$U = q\bigl(f_1^{-1}(U_1 \times [0,1)) \cup f_2^{-1}(U_2 \times [0,1))\bigr) $$ The map $\psi : U \to B \times (-1,+1)$ will be given by the following formula. Each $y \in U$ has one of two forms, and we give the formula for each form:

  • If $y = qf_1^{-1}(z,t)$ for $(z,t) \in B \times [0,1)$ then $\psi(y) = (z,-t) \in B \times (-1,0]$
  • If $y = q f_2^{-1}(z,t)$ for $(z,t) \in B \times [0,1)$ then $\psi(y) = (z,t)$

This is well-defined if $y$ has both forms (which only happens when $t=0$).

And then, by tracing through all the definitions and using all the theorems you can possibly find from quotient spaces, it follows that $\psi$ is a homeomorphism from $U$ to $B \times (-1,+1)$.

$\endgroup$

Your Answer

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