Graduate Connected sum of manifolds and free group isomorphisms

[PsychonautQQ]

Let $M$ and $N$ be connected n-manifolds, n>2. Prove that the fundamental group of $M#N$ (the connected sum of $M$ and $N$) is isomorphic to $\pi(M)* \pi(N)$ (the free group of the fundamental groups of $M$ and $N$)

This is not for homework, I was hoping to get some insight here. Can anyone help me get started on how to show this?

 

[lavinia]

Let $M$ and $N$ be connected n-manifolds, n>2. Prove that the fundamental group of $M#N$ (the connected sum of $M$ and $N$) is isomorphic to $\pi(M)* \pi(N)$ (the free group of the fundamental groups of $M$ and $N$)

This is not for homework, I was hoping to get some insight here. Can anyone help me get started on how to show this?

Did you try Van Kampen's Theorem?

 

[lavinia]

Let $M$ and $N$ be connected n-manifolds, n>2. Prove that the fundamental group of $M#N$ (the connected sum of $M$ and $N$) is isomorphic to $\pi(M)* \pi(N)$ (the free group of the fundamental groups of $M$ and $N$)

This is not for homework, I was hoping to get some insight here. Can anyone help me get started on how to show this?

@PsychonautQQ I would be interested to see your proof.

$n>2$ is important,

 

[WWGD]

Let $M$ and $N$ be connected n-manifolds, n>2. Prove that the fundamental group of $M#N$ (the connected sum of $M$ and $N$) is isomorphic to $\pi(M)* \pi(N)$ (the free group of the fundamental groups of $M$ and $N$)

Don't you mean free product of the two groups? I would suggest tinker with familiar and extreme cases: trivial groups on one , both, and known cases, for motivation/context.

 

[PsychonautQQ]

@PsychonautQQ I would be interested to see your proof.

$n>2$ is important,

$n>2$ is what makes the intersection of the two open sets simply connected. I am trying to think of a way to pick an open set for $M$ such that it 'reaches' into $N$ a little bit and is still open and also can deformation retract back into $M$. That way the fundamental group of this open set will still be $\pi(M)$ and if I find a similar neighborhood that 'reaches' into $M$ and contains $N$ then I think I will be good.

 

[lavinia]

$n>2$ is what makes the intersection of the two open sets simply connected. I am trying to think of a way to pick an open set for $M$ such that it 'reaches' into $N$ a little bit and is still open and also can deformation retract back into $M$. That way the fundamental group of this open set will still be $\pi(M)$ and if I find a similar neighborhood that 'reaches' into $M$ and contains $N$ then I think I will be good.

Right.

Notice that when you do the deformation retractions you will retract onto $M$ minus an open ball and $N$ minus an open ball.

 

[PsychonautQQ]

Right.

Notice that when you do the deformation retractions you will retract onto $M$ minus an open ball and $N$ minus an open ball.

and $\pi(M)$ is isomorphic to $\pi(M-b)$ where b is a euclidean ball (for $n>2$).

But how do I choose an open neighborhood $U$ such that $M$ reaches into $N$ a bit?

 

[lavinia]

and $\pi(M)$ is isomorphic to $\pi(M-b)$ where b is a euclidean ball (for $n>2$).

But how do I choose an open neighborhood $U$ such that $M$ reaches into $N$ a bit?

How did you choose it for a manifold minus an open ball?

 

[PsychonautQQ]

How did you choose it for a manifold minus an open ball?

The whole manifold minus the boundary of the open ball? I'm not sure actually.

 

[lavinia]

The whole manifold minus the boundary of the open ball? I'm not sure actually.

Try the example of Euclidean space with the standard Euclidean metric minus an open ball.

 

[PsychonautQQ]

The Euclidean space minus an open ball is an open set.
So in the case of Manifolds we can take a manifold minus a neighborhood homeomorphic to a euclidean ball and it will still be open. Therefore when 'reaching' into $N$ we can just union an open Euclidean ball in $N$ with $M$ and it will be an open set.

 

[lavinia]

The Euclidean space minus an open ball is an open set.

Euclidean space minus an open ball is a closed set.

So in the case of Manifolds we can take a manifold minus a neighborhood homeomorphic to a euclidean ball and it will still be open. Therefore when 'reaching' into $N$ we can just union an open Euclidean ball in $N$ with $M$ and it will be an open set.

I don't see why there would be an open ball in $N$ that extends the manifold $M$ into $N$.

 

[PsychonautQQ]

Then how do I use Seifert Van-Kampen's theorem?

 

[lavinia]

Then how do I use Seifert Van-Kampen's theorem?

Without giving it away, there will be a different open set that extends $N$ into $M$.

 

[PsychonautQQ]

N union an open set contained within the boundary of the disk we remove from M when forming the connected sum?

 

[lavinia]

N union an open set contained within the boundary of the disk we remove from M when forming the connected sum?

Can you show me an open set in $M$ that is contained in the boundary of the removed disk?