Genus-g Surface and Retraction to Circle

[Bacle]

Hi, All:

I saw an argument in another site re the claim that the genus-g surface Sg does not retract to a circle. The argument was that,using/assuming H_1(Sg,Z)=Z^{2g};
and H_1(C,Z)=Z ; Z the integers and H_1(Sg,Z) if there was a retraction r: Sg-->C , then , for i being the inclusion ; r the retraction, then roi would give an isomorphism Z-->Z , which cannot happen with a composition Z-->Z^{2g}.

The counter ** is that we can compose Z-->Z^{2g}-->Z by n-->(0,..,0,n,..)-->n
by injection i_k into k-th component composed with projection pi_k into k-th component.

I think the counter has no valid grounds; but I cannot think of a good argument against
the counter, other than that the composition of non-isomorphism cannot be an isomorphism, since the automorphisms form a group .

Can anyone think of any better counter to the counter ** ?

Let f: Z-->Z be a group isomorphism , then , use that automorphism

 

[lavinia]

Hi, All:

I saw an argument in another site re the claim that the genus-g surface Sg does not retract to a circle. The argument was that,using/assuming H_1(Sg,Z)=Z^{2g};
and H_1(C,Z)=Z ; Z the integers and H_1(Sg,Z) if there was a retraction r: Sg-->C , then , for i being the inclusion ; r the retraction, then roi would give an isomorphism Z-->Z , which cannot happen with a composition Z-->Z^{2g}.

The counter ** is that we can compose Z-->Z^{2g}-->Z by n-->(0,..,0,n,..)-->n
by injection i_k into k-th component composed with projection pi_k into k-th component.

I think the counter has no valid grounds; but I cannot think of a good argument against
the counter, other than that the composition of non-isomorphism cannot be an isomorphism, since the automorphisms form a group .

Can anyone think of any better counter to the counter ** ?

Let f: Z-->Z be a group isomorphism , then , use that automorphism

I think the counter is correct. This argument should be tried on the fundamental groups.

 

[Bacle]

But, can't we just use the fact that a retraction from Sg to C would give us

an isomorphism from Z to Z+Z ? Also, can the composition of maps that are not

isomorphisms be an isomorphisms; can we get a composition Z-->Z(+)Z-->Z be

an isomorphism if neither of the maps is an isomorphism?

 

[lavinia]

But, can't we just use the fact that a retraction from Sg to C would give us

an isomorphism from Z to Z+Z ? Also, can the composition of maps that are not

isomorphisms be an isomorphisms; can we get a composition Z-->Z(+)Z-->Z be

an isomorphism if neither of the maps is an isomorphism?

Inclusion into the first factor followed by projection onto the first factor

 

[Bacle]

Yes; these are exactly the maps I referred-to in my 2nd post; I guess a composition of
Algebraic embeddings can be a homomorphism; I imagine the same can happen in other "categories".

 

[lavinia]

Yes; these are exactly the maps I referred-to in my 2nd post; I guess a composition of
Algebraic embeddings can be a homomorphism; I imagine the same can happen in other "categories".

like homeomorphism or diffeomorphism etc

 

[Bacle]

But the original question was interesting: we get an Sg; genus-g orientable surface,

with g>1 , in which we draw two circles C and C' embedded in Sg , so that C separates Sg -- meaning that

Sg-C is disconnected-- but C' does not separate in this same sense. Then we're asked to show

that Sg retracts to C', but does not retract to C.

The confusion is in trying to use the argument that, if a retraction r: Sg-->C existed, then

we would have isomorphisms between the respective Pi_1's, or the respective H_1's

(tho we could also use the respectives H_2 to show there is no iso.). Fine, then,

but a problem, then, is: why can we use this argument for the retractions works

for a retraction r:Sg-->C , why doesn't it work for a retraction r': Sg-->C' ?

 

[lavinia]

But the original question was interesting: we get an Sg; genus-g orientable surface,

with g>1 , in which we draw two circles C and C' embedded in Sg , so that C separates Sg -- meaning that

Sg-C is disconnected-- but C' does not separate in this same sense. Then we're asked to show

that Sg retracts to C', but does not retract to C.

The confusion is in trying to use the argument that, if a retraction r: Sg-->C existed, then

we would have isomorphisms between the respective Pi_1's, or the respective H_1's

(tho we could also use the respectives H_2 to show there is no iso.). Fine, then,

but a problem, then, is: why can we use this argument for the retractions works

for a retraction r:Sg-->C , why doesn't it work for a retraction r': Sg-->C' ?

I am having trouble seeing the construction. Do I cut along C' and take one of the pieces then cut a second time?

 

[Bacle]

Sorry, but this is exercise 9 in p.53 of Hatcher:

http://www.math.cornell.edu/~hatcher/AT/AT.pdf

I don't know how to scan it so it is shown in here.

 

[lavinia]

Sorry, but this is exercise 9 in p.53 of Hatcher:

http://www.math.cornell.edu/~hatcher/AT/AT.pdf

I don't know how to scan it so it is shown in here.

I looked ay Hatcher. The two half surfaces do not retract onto the circle,C,because C is a homology boundary. That is the meaning of his hint about abelianizaing the fundamental group.

 

[WWGD]

Please forgive my ignorance, Lavinia, but why can't a boundary be a retract of a space?

 

[lavinia]

Please forgive my ignorance, Lavinia, but why can't a boundary be a retract of a space?

In this case the boundary is a circle. It is a homology boundary so the inclusion map of this circle induces the map Z -> 0 on homology. If it were a retract then the composition of the group homomorphisms on homology, Z -> 0 -> Z, would have to be the identity map. This is the same argument as in the classic proof of the Brouwer Fixed point theorem for the 2 disk. More generally though, an oriented compact manifold has top homology isomorphic to Z. If it is the boundary of another manifold, then it is homologous to zero and can not be a retract of that manifold. So a sphere is not a retract of a ball. A torus is not a retract of a solid torus.

 

[WWGD]

I see, it makes sense. But if the circle was non-bounding , then its homology would
be that of S^1, so that , for this case, the argument , as you and Bacle discussed, does not necessarily work, since you may compose non-homeomorphsims from Z to Z(+)Z to Z to get an isomorphism, right?

Still, above argument clearly does not imply that a retraction of Sg into the non-boundary is possible. How would one argue then, that the retraction does exist?

 

[lavinia]

I see, it makes sense. But if the circle was non-bounding , then its homology would
be that of S^1, so that , for this case, the argument , as you and Bacle discussed, does not necessarily work, since you may compose non-homeomorphsims from Z to Z(+)Z to Z to get an isomorphism, right?

Still, above argument clearly does not imply that a retraction of Sg into the non-boundary is possible. How would one argue then, that the retraction does exist?

right. Then the argument wouldn't work. But in Hatcher's example he splits the manifold along a circle that is not null homologous into two manifolds with boundary then says well it can't be a retract on the pieces so it can not be a retract on the whole. The second circle does not separate the surface into two disjount pieces, Why doesn't the argument work then?

 

[lavinia]

- My statement about an oriented manifold being a homology boundary when it is the boundary of another manifold assumes that the other manifold is orientable. A counter example would be the Moebius band, a non-orientable manifold, whose boundary is a circle. This circle is not null homologous. However it is double a generator so the same argument applies. I wonder what happens when the bounding manifold is not orientable.

- A specific case of the second part of Hatcher's problem is the torus. Here the circle that cuts transversally does not separate the torus into two pieces but rather into a cylinder. This circle is a retract to the torus.

 

[Jamma]

I'm having difficulty following this: surely the circle is never a retract of any closed surface simply by looking at the homology groups, no matter what the circle does, whether it is null homologous, splits the surface into connected components...

 

[Jamma]

I'm having difficulty following this: surely the circle is never a retract of any closed surface simply by looking at the homology groups, no matter what the circle does, whether it is null homologous, splits the surface into connected components...

 

[lavinia]

I'm having difficulty following this: surely the circle is never a retract of any closed surface simply by looking at the homology groups, no matter what the circle does, whether it is null homologous, splits the surface into connected components...

The torus is the circle Cartesian product the circle. Project onto the first factor.

 

[Jamma]

Oh, of course Was thinking about the maps the wrong way round!

So the theorem is that it is a retract if and only if it doesn't separate the space into two pieces?

 

[lavinia]

Oh, of course Was thinking about the maps the wrong way round!

So the theorem is that it is a retract if and only if it doesn't separate the space into two pieces?

That seems like the hypothesis but I don't see the whole proof yet.

 

[Jamma]

Perhaps you can just use the classification of 2-manifolds?

In one direction, I think that we have shown that if the circle is a boundary then we don't have a retract.

However, in the other direction we have that the 2-manifold is a connected sum of tori. So our circle is homologous to something specific where we can say that the retract will just be the analogous version of "project to the n'th coordinate".

 

[lavinia]

Perhaps you can just use the classification of 2-manifolds?

In one direction, I think that we have shown that if the circle is a boundary then we don't have a retract.

However, in the other direction we have that the 2-manifold is a connected sum of tori. So our circle is homologous to something specific where we can say that the retract will just be the analogous version of "project to the n'th coordinate".

you would think the argument would go something like that... but not sure