Showing that $\pi(M \# N) = \pi(M) \ast \pi(N)$ for $n$-dimensional manifolds $M$,$N$

Problem: Let $M$ and $N$ be $n$-dimensional manifolds, where $n > 2$. Let $M \# N$ be their connected sum. Show that $\pi(M \# N) = \pi(M) \ast \pi(N)$.

RE-EDITED Attempt:

1. Let $U_2$ and $V_2$ be two small open balls from $M$ and $N$ to be removed by the connected sum operation. Let $p_m \in U_2$ and $p_n \in V_2$, and then let $U_1 = M - \{p_m\}$ and $U_2 = N - \{p_n\}$.

2. We can now view the connected sum $M\#N$ as the quotient of the disjoint union of $M$ and $N$ by an equivalence relation identifying $U_2 -\{p_m\}$ with $V_2-\{p_n\}$ defined by some homeomorphism between the two. Denote $W_1$ and $W_2$ as the respective images of $U_1$ and $V_1$ under the quotient map.

3. We have that $U_2$ and $V_2$ are open by construction. Furthermore, since $M$ and $N$ are open, we have that $U_1 = M - \{p_m\}$ and $V_1 = N - \{p_n\}$ are also open (since open sets with a point removed are still open).

4. Then we have the following open covers of $M$ and $N$ respectively:

$$M = U_1 \cup U_2$$ $$N = V_1 \cup V_2$$

5. We can then express

$$M \# N = \underbrace{W_1}_{\text{open}} \cup \underbrace{W_2}_{\text{open}}$$

as well since $M \cup N = V_1 \cup U_1$ and $W_1$ and $W_2$ are just the images of $U_1$ and $V_1$ under the natural quotient map used to define $M \# N$.

6. Consider that $W_1 \cap W_2$ is path connected since $W_1 \cap W_2$ is homeomorphic to $U_2 - \{p_m\} \cong V_2 - \{p_n\}$, both of which are punctured disks which deformation retract onto $S^{n-1}$. Since spheres of dimension greater than $2$ are simply connected (hence path connected), we have that $W_1 \cap W_2$ is path connected as well. This will allow us to later apply Van Kampen to $M \# N = W_1 \cup W_2$.

7. Now we have that

$$U_1 \cap U_2 = U_2 - \{p_n\} \cong W_1 \cap W_2 \cong V_2 - \{p_m\} = V_1 \cap V_2$$

so that from above we can say

$$\pi_1(U_1 \cap U_2) \cong \pi(V_1 \cap V_2) \cong \pi(W_1 \cap W_2) \cong \{e\}$$

8. Now since $n > 2$, we have that

$$\underbrace{\pi_1(U_1) = \pi_1(M - \{p_m\}) \cong \pi_1(M) \cong \pi_1(W_1)}_{\text{removing a point doesn't change fundamental group for n > 2}}$$

and similarly

$$\pi_1(V_1) = \pi_1(N - \{p_n\})\cong\pi_1(N) \cong \pi_1(W_2)$$

9. Then applying Van Kampen on $M \# N = W_1 \cup W_2$ yields that

$$\pi_1(W_1) \ast_{\pi_1(W_1 \cap W_2)} \pi_1(W_2) \cong \pi_1(M\#N)$$

10. But since (8) yields that

$$\pi_1(W_1) \ast_{\pi_1(W_1 \cap W_2)} \pi_1(W_2) \cong \pi_1(W_1) \ast_{\{e\}} \pi_1(W_2) \cong \pi_1(W_1) \ast \pi_1(W_2) \cong \pi_1(M) \ast \pi_1(N)$$

we then have from (9) that

$$\pi_1(M) \ast \pi_1(N) \cong \pi_1(M\#N)$$

as desired.

• Hint: use Seifert–van Kampen – Grigory M May 9 '14 at 10:55
• $\pi_1(M)$ is a group not a number, so «$\pi_1(M)=\infty$» is not even wrong – Grigory M May 9 '14 at 10:56
• @user1770201: What is $\pi$? – user99914 May 9 '14 at 11:07
• @John: The fundamental group, or the set of all paths (up to homotopy) of $M$ or $N$. Since $M$ and $N$ are manifolds, I presume our choice of base points don't matter. – user1770201 May 9 '14 at 11:11
• In light of the duplicate post flagged by Grigory M and other users, I edited my post above with (what I think is) a complete solution to the problem but with questions flagged on the steps where I'm specifically confused. I appreciate the comments. – user1770201 May 9 '14 at 15:28

1 Answer

Let's be a little more precise. The direct sum $M\#N$ is defined as quotient of the disjoint union of $M$ and $N$ by an equivalence relation identifying $U_2 -\{p_m\}$ with $V_2-\{p_n\}$ defined by some homeomorphism between the two. Then denote by $W_1$ the image of $U_1$ under the quotient map and by $W_2$ the image of $V_1$ under the quotient map.

5) $W_1\cap W_2$ is connected because it is homeomorphic to $U_2-\{p_m\}\cong V_2-\{p_n\}$. These are punctured disks, which deformation retract onto spheres. Spheres of dimension greater than $0$ are connected.

(Note that this still works for $n=2$.)

6) See above: everything here intersects in the punctured disk where we glued to create the connect sum. Punctured disks deformation retract to spheres.

7) The space $S^{n-1}$ is definitely not contractible for $n\geq 2$. It is simply connected.

8) Here is where you need the hypothesis $n>2$. For dimension greater than $2$, removing a point does not change the fundamental group. To see why, show that the homomorphism $\pi_1(U_1)\to \pi_1(M)$ induced by the inclusion is actually an isomorphism. The intuition is that if any of the homotopies between generators $\pi_1(M)$ hits the ball you've removed, it can be "moved" so it "goes around" the ball.

(Removing a point does change higher homotopy groups --- $M$ is not homotopy equivalent to $U_1$ --- but does not change $\pi_1$.)

• In your first paragraph, why is the equivalence relation identifying $U_2 - \{p_m\}$ with $V_2 - \{p_n\}$ instead of just $U_2$ with $V_2$? Also in your first paragraph: how do we know what the image of $U_1$ and $V_1$ is under the quotient map for purposes of defining $W_1$ and $W_2$? – user1770201 May 11 '14 at 16:58
• @user1770201 Because $U_2$ and $V_2$ are open disk; for a connect sum don't want to glue the two manifolds along a disk, we want to glue them along a collar of the removed disks. – Neal May 11 '14 at 17:24
• @user1770201 I'm not sure I understand the question. The image of $U_1$ is the set of equivalence classes in $M\# N$ which have at least one representative in $U_1$. – Neal May 11 '14 at 17:26
• Do you have a link which explains in more detail the definition of $M \# N$ you are using? Definitions seem to vary from site to site (wikipedia, proof wiki, etc) and I'm trying to find yours. – user1770201 May 12 '14 at 2:21
• @user1770201 They're all the same: Delete a disk, glue along boundaries. :) Mine is equivalent to Wikipedia's in the case of a smooth manifold; the disk $U_i$ is a tubular neighborhood of $p_i$ diffeomorphic to the normal bundle of $p_i$. Wikipedia also specifies a choice of gluing map, which I did not do. – Neal May 12 '14 at 2:36