This is problem 4.19 in Lee's Topological Manifolds. I have been working on this problem for a couple days now, and just a need a hint in the right direction.

If $M$ and $N$ are two $n$-manifolds, and $B_1\subset M$ and $B_2\subset N$ are two open, regular coordinate balls (definition below), the connected sum $M\#N$ is the quotient of the disjoint union $(M-B_1)\displaystyle\sqcup (N-B_2)$ by the relation that identifies points on the spherical boundaries of each component via some homeomorphism $h$.

Now I want to show that there are two open sets $U,V\subset M\#N$, such that:

  1. $U\cong M-\{p\}$ and $V\cong N-\{q\}$, for some points $p\in M$ and $q\in N$
  2. $U\cap V\cong S^{n-1}\times\mathbb{R}$
  3. $U\cup V=M\#N$

Here's my intuition: take a larger coordinate ball $D$ in $N$, containing $B_2$, which works because $B_2$ is regular. $(M-B_1)\sqcup (D-B_2)$ is a saturated open set, so its image in $M\#N$ is open. This is what I want to be $U$.

Now I've managed to define a homeomorphism $g$ from $D-B_2$ to $\overline{\mathbb{B}}_1(0)-\{0\}$ (the punctured closed unit ball). If I can then map that punctured unit ball to $\overline{B}_1-\{p\}$, I can try and show the map $f:\ (M-B_1)\sqcup(D-B_2)\rightarrow M-\{p\}$

$$ f(x)=\begin{cases} x& x\in M-B_1\\ g(x)& x\in D-B_2\\ \end{cases} $$

is a quotient map, and then use the uniqueness of quotients to show $U$ is homeomorphic to $M-\{p\}$. However, this is too hard for me to do, because I need to somehow incorporate $h$ in the mix, so that the identifications of $f$ match those of $h$.

Can someone provide a hint here? I feel like this is a lot of work for one problem, and I understand intuitively what to do here, but getting all the details right is proving to be too much.

I'm also open to hints about part 2. as well (3. is easy).

Definition: A coordinate ball $B\subset M$ is called regular if it has a neighborhood $B'\supset B$, such that there is a homeomorphism $k: B'\rightarrow \mathbb{B}_s(0)$ (this is the open ball of radius $s$ around $0$). Under this homeomorphism, $B$ goes to $\mathbb{B}_r(0)$ for some $0<r<s$, and $\overline{B}$ goes to $\overline{\mathbb{B}}_r(0)$.

  • 2
    $\begingroup$ You could make your life easy by actually choosing $h$ to satisfy some extremely simple formula. That should make it easy to make the identifications of $f$ and $h$ match. $\endgroup$
    – Lee Mosher
    Jan 27, 2018 at 0:36

1 Answer 1


I agree this is a tedious problem, but I think you have basically got it. I'll try and use your terminology throughout. I'll assume the homeomorphism you mentioned is $h: \delta B_2\rightarrow\delta B_1$.

Now you have a homeomorphism $g: D-B_2\rightarrow \overline{\mathbb{B}}_t(0)-\{0\}$ (you used a radius of $1$, but I'm using a radius of $t$, which is just a difference of a little scaling). I'm going to assume we also have a homeomorphism $k: \overline{B_1}\rightarrow \overline{\mathbb{B}}_t(0)$, as in your definition.

As you mentioned, $k^{-1}\circ g$ is a homeomorphism from $D-B_2$ to $\overline{B}_1-\{p\}$ (assuming $k(p)=0$). The problem (as you also mentioned) is what happens on the boundary: you need it to do exactly what $h$ does. The key is that we can do something between $g$ and $k^{-1}$ that helps with that.

Thinking just about the boundary, we want to find a map $r: \delta\mathbb{B}_t(0)\rightarrow \delta\mathbb{B}_t(0)$, such that $$ k^{-1}\circ r\circ g\equiv h$$

as maps from $\delta B_2$ to $\delta B_1$.

Since everything in sight is a homeomorphism, there's really only one choice for $r$: $r=k\circ h\circ g^{-1}$. But we need this to be defined on all of $\overline{\mathbb{B}}_t(0)-\{0\}$.

If we look at the punctured ball as just a union of a bunch of spheres of varying (positive!) radii, the extension seems pretty natural. We define $\tilde{r}: \overline{\mathbb{B}}_t(0)-\{0\}\rightarrow\overline{\mathbb{B}}_t(0)-\{0\}$ as $$ \tilde{r}(x)=|x|r\left(\frac{x}{|x|}\right) $$

It's easy to check this is a homeomorphism (the inverse is just $|x|r^{-1}\left(\frac{x}{|x|}\right)$). Now, if we (finally!) define $G=k^{-1}\circ\tilde{r}\circ g$, we have a homeomorphism from $D-B_2$ to $\overline{B}_1-\{p\}$. And more importantly,

$$ G(x) = h(x)\text{ for }x\in\delta B_2 $$

So we can define $f: (M-B_1)\sqcup(D-B_2)\rightarrow M-\{p\}$ as $$ f(x)=\begin{cases} x& x\in M-B_1\\ G(x)& x\in D-B_2\\ \end{cases} $$

This map is continuous and surjective. You mentioned showing it is a quotient map, and that's pretty easy, since each "branch" is a homeomorphism. And with the help of $\tilde{r}$, this map now makes the same identifications as the original quotient map.

As for part (2), I'll just give a hint: it helps to look at the preimage of $U\cap V$ under the quotient map. You already know what the "$N$" part looks like ($D-B_2$). Perhaps there's a nice quotient map, from the "$N$" and "$M$" pieces, to $\mathbb{S}^{n-1}\times\mathbb{R}$?

  • $\begingroup$ Thanks! I've spent the morning working through your answer, and now feel I've fully "grokked" it. $\endgroup$ Jan 27, 2018 at 15:50
  • $\begingroup$ @SteveD : This is nice anwer. I'm working on this problem too. If we allow to choose $h$, can we just set $h := (k^{-1} \circ g)|_{\partial B_2}$ (as LeeMosher said) ? $\endgroup$ Apr 28, 2018 at 11:08
  • $\begingroup$ @Sou: yeah I think so. But where is the fun in that? :) $\endgroup$
    – Steve D
    Apr 28, 2018 at 13:19
  • $\begingroup$ @Sou: Well, the problem statement doesn't explicitly give you the option to choose $h$, so I answered the problem the OP asked: where $h$ is already given, and you have to work around it. I agree it is simpler to choose $h$ yourself. $\endgroup$
    – Steve D
    Apr 28, 2018 at 15:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .