On the connected sum of a surface with a torusSphere with three Möbius strips glued and sphere with a handle...
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On the connected sum of a surface with a torus
Sphere with three Möbius strips glued and sphere with a handle and a Möbius strip glued3-fold connected covers of punctured torusClosed, orientable surface whose genus is very hard to find intuitivelyConnected Sum of SurfacesOrbit space of torus homeomorphic to mobius stripThe sphere as a closed combinatorial surfaceGenus of an Orientable Surface Must Be an IntegerWhat shape is homeomorphic to a sphere attached with a handle and a cross cap?Connected sum of two non homeomorphic surfacesFundamental polygon of surface with boundary
$begingroup$
I am studying the classification of Surfaces, and run into the notion of connected sum. We define it in terms of triangulations. I want to show the following. Let $S$ be a triangulated surface. I want to show that $S#T$ with $T$ the torus is combinatorially homeomorphic to attaching a handle to $S$. I am not allowed to use the classification of Surfaces, since this is what I want to prove.
So, Let $S'$ denote the complement of a combinatorial disk in $S$, Then I want to glue the boundary of $S'$ to the complement of a combinatorial disk in $T$. How is this combinatorially homeomorphic to $S$ with a handle glued in?
EDIT: Possible solution: Since $S#S^2cong S$, and $T$ is $S^2$ with a handle glued in, $S#Tcong S#(S^2text{with handle})cong (S# S^2)text{with a handle glued in}$. Is this last line correct?
general-topology algebraic-topology surfaces geometric-topology
$endgroup$
add a comment |
$begingroup$
I am studying the classification of Surfaces, and run into the notion of connected sum. We define it in terms of triangulations. I want to show the following. Let $S$ be a triangulated surface. I want to show that $S#T$ with $T$ the torus is combinatorially homeomorphic to attaching a handle to $S$. I am not allowed to use the classification of Surfaces, since this is what I want to prove.
So, Let $S'$ denote the complement of a combinatorial disk in $S$, Then I want to glue the boundary of $S'$ to the complement of a combinatorial disk in $T$. How is this combinatorially homeomorphic to $S$ with a handle glued in?
EDIT: Possible solution: Since $S#S^2cong S$, and $T$ is $S^2$ with a handle glued in, $S#Tcong S#(S^2text{with handle})cong (S# S^2)text{with a handle glued in}$. Is this last line correct?
general-topology algebraic-topology surfaces geometric-topology
$endgroup$
$begingroup$
It's hard to answer your question because you have not given us your definition of "$S$ with a handle glued in". In some treatments, the definition literally is "$S # T$", in which case the answer to your question would be "this is true by definition". If that's not a satisfactory answer to you, and I suspect it is not, then you need to fill your question out by telling us exactly what you mean by gluing a handle in.
$endgroup$
– Lee Mosher
Mar 20 at 15:11
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We have defined gluing in a handle as follows: we choose two triangles in a triangulation which are disjoint, and remove the interiors. Then we identify the resulting edges
$endgroup$
– user408856
Mar 20 at 15:42
$begingroup$
Then I am wondering, are the following operations equivalent: $S#(S^2text{with one handle glued in })$ and $(S#S^2)$ with a handle glued in
$endgroup$
– user408856
Mar 20 at 15:45
$begingroup$
Does this follow from the disk theorem?
$endgroup$
– user408856
Mar 20 at 15:53
$begingroup$
There's a problem in that you still have not said carefully enough what you mean by "we identify the resulting edges". If you do not do this very carefully, there are a lot of other possible results. For example, instead of getting $S#$ (torus) you can also get $S#$(Klein bottle) (this corrects an error in the previous version of this comment). You should add to the question itself (not using a comment) exactly what you mean by gluing in a handle.
$endgroup$
– Lee Mosher
Mar 21 at 0:45
add a comment |
$begingroup$
I am studying the classification of Surfaces, and run into the notion of connected sum. We define it in terms of triangulations. I want to show the following. Let $S$ be a triangulated surface. I want to show that $S#T$ with $T$ the torus is combinatorially homeomorphic to attaching a handle to $S$. I am not allowed to use the classification of Surfaces, since this is what I want to prove.
So, Let $S'$ denote the complement of a combinatorial disk in $S$, Then I want to glue the boundary of $S'$ to the complement of a combinatorial disk in $T$. How is this combinatorially homeomorphic to $S$ with a handle glued in?
EDIT: Possible solution: Since $S#S^2cong S$, and $T$ is $S^2$ with a handle glued in, $S#Tcong S#(S^2text{with handle})cong (S# S^2)text{with a handle glued in}$. Is this last line correct?
general-topology algebraic-topology surfaces geometric-topology
$endgroup$
I am studying the classification of Surfaces, and run into the notion of connected sum. We define it in terms of triangulations. I want to show the following. Let $S$ be a triangulated surface. I want to show that $S#T$ with $T$ the torus is combinatorially homeomorphic to attaching a handle to $S$. I am not allowed to use the classification of Surfaces, since this is what I want to prove.
So, Let $S'$ denote the complement of a combinatorial disk in $S$, Then I want to glue the boundary of $S'$ to the complement of a combinatorial disk in $T$. How is this combinatorially homeomorphic to $S$ with a handle glued in?
EDIT: Possible solution: Since $S#S^2cong S$, and $T$ is $S^2$ with a handle glued in, $S#Tcong S#(S^2text{with handle})cong (S# S^2)text{with a handle glued in}$. Is this last line correct?
general-topology algebraic-topology surfaces geometric-topology
general-topology algebraic-topology surfaces geometric-topology
edited Mar 20 at 15:50
asked Mar 20 at 12:35
user408856
$begingroup$
It's hard to answer your question because you have not given us your definition of "$S$ with a handle glued in". In some treatments, the definition literally is "$S # T$", in which case the answer to your question would be "this is true by definition". If that's not a satisfactory answer to you, and I suspect it is not, then you need to fill your question out by telling us exactly what you mean by gluing a handle in.
$endgroup$
– Lee Mosher
Mar 20 at 15:11
$begingroup$
We have defined gluing in a handle as follows: we choose two triangles in a triangulation which are disjoint, and remove the interiors. Then we identify the resulting edges
$endgroup$
– user408856
Mar 20 at 15:42
$begingroup$
Then I am wondering, are the following operations equivalent: $S#(S^2text{with one handle glued in })$ and $(S#S^2)$ with a handle glued in
$endgroup$
– user408856
Mar 20 at 15:45
$begingroup$
Does this follow from the disk theorem?
$endgroup$
– user408856
Mar 20 at 15:53
$begingroup$
There's a problem in that you still have not said carefully enough what you mean by "we identify the resulting edges". If you do not do this very carefully, there are a lot of other possible results. For example, instead of getting $S#$ (torus) you can also get $S#$(Klein bottle) (this corrects an error in the previous version of this comment). You should add to the question itself (not using a comment) exactly what you mean by gluing in a handle.
$endgroup$
– Lee Mosher
Mar 21 at 0:45
add a comment |
$begingroup$
It's hard to answer your question because you have not given us your definition of "$S$ with a handle glued in". In some treatments, the definition literally is "$S # T$", in which case the answer to your question would be "this is true by definition". If that's not a satisfactory answer to you, and I suspect it is not, then you need to fill your question out by telling us exactly what you mean by gluing a handle in.
$endgroup$
– Lee Mosher
Mar 20 at 15:11
$begingroup$
We have defined gluing in a handle as follows: we choose two triangles in a triangulation which are disjoint, and remove the interiors. Then we identify the resulting edges
$endgroup$
– user408856
Mar 20 at 15:42
$begingroup$
Then I am wondering, are the following operations equivalent: $S#(S^2text{with one handle glued in })$ and $(S#S^2)$ with a handle glued in
$endgroup$
– user408856
Mar 20 at 15:45
$begingroup$
Does this follow from the disk theorem?
$endgroup$
– user408856
Mar 20 at 15:53
$begingroup$
There's a problem in that you still have not said carefully enough what you mean by "we identify the resulting edges". If you do not do this very carefully, there are a lot of other possible results. For example, instead of getting $S#$ (torus) you can also get $S#$(Klein bottle) (this corrects an error in the previous version of this comment). You should add to the question itself (not using a comment) exactly what you mean by gluing in a handle.
$endgroup$
– Lee Mosher
Mar 21 at 0:45
$begingroup$
It's hard to answer your question because you have not given us your definition of "$S$ with a handle glued in". In some treatments, the definition literally is "$S # T$", in which case the answer to your question would be "this is true by definition". If that's not a satisfactory answer to you, and I suspect it is not, then you need to fill your question out by telling us exactly what you mean by gluing a handle in.
$endgroup$
– Lee Mosher
Mar 20 at 15:11
$begingroup$
It's hard to answer your question because you have not given us your definition of "$S$ with a handle glued in". In some treatments, the definition literally is "$S # T$", in which case the answer to your question would be "this is true by definition". If that's not a satisfactory answer to you, and I suspect it is not, then you need to fill your question out by telling us exactly what you mean by gluing a handle in.
$endgroup$
– Lee Mosher
Mar 20 at 15:11
$begingroup$
We have defined gluing in a handle as follows: we choose two triangles in a triangulation which are disjoint, and remove the interiors. Then we identify the resulting edges
$endgroup$
– user408856
Mar 20 at 15:42
$begingroup$
We have defined gluing in a handle as follows: we choose two triangles in a triangulation which are disjoint, and remove the interiors. Then we identify the resulting edges
$endgroup$
– user408856
Mar 20 at 15:42
$begingroup$
Then I am wondering, are the following operations equivalent: $S#(S^2text{with one handle glued in })$ and $(S#S^2)$ with a handle glued in
$endgroup$
– user408856
Mar 20 at 15:45
$begingroup$
Then I am wondering, are the following operations equivalent: $S#(S^2text{with one handle glued in })$ and $(S#S^2)$ with a handle glued in
$endgroup$
– user408856
Mar 20 at 15:45
$begingroup$
Does this follow from the disk theorem?
$endgroup$
– user408856
Mar 20 at 15:53
$begingroup$
Does this follow from the disk theorem?
$endgroup$
– user408856
Mar 20 at 15:53
$begingroup$
There's a problem in that you still have not said carefully enough what you mean by "we identify the resulting edges". If you do not do this very carefully, there are a lot of other possible results. For example, instead of getting $S#$ (torus) you can also get $S#$(Klein bottle) (this corrects an error in the previous version of this comment). You should add to the question itself (not using a comment) exactly what you mean by gluing in a handle.
$endgroup$
– Lee Mosher
Mar 21 at 0:45
$begingroup$
There's a problem in that you still have not said carefully enough what you mean by "we identify the resulting edges". If you do not do this very carefully, there are a lot of other possible results. For example, instead of getting $S#$ (torus) you can also get $S#$(Klein bottle) (this corrects an error in the previous version of this comment). You should add to the question itself (not using a comment) exactly what you mean by gluing in a handle.
$endgroup$
– Lee Mosher
Mar 21 at 0:45
add a comment |
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$begingroup$
It's hard to answer your question because you have not given us your definition of "$S$ with a handle glued in". In some treatments, the definition literally is "$S # T$", in which case the answer to your question would be "this is true by definition". If that's not a satisfactory answer to you, and I suspect it is not, then you need to fill your question out by telling us exactly what you mean by gluing a handle in.
$endgroup$
– Lee Mosher
Mar 20 at 15:11
$begingroup$
We have defined gluing in a handle as follows: we choose two triangles in a triangulation which are disjoint, and remove the interiors. Then we identify the resulting edges
$endgroup$
– user408856
Mar 20 at 15:42
$begingroup$
Then I am wondering, are the following operations equivalent: $S#(S^2text{with one handle glued in })$ and $(S#S^2)$ with a handle glued in
$endgroup$
– user408856
Mar 20 at 15:45
$begingroup$
Does this follow from the disk theorem?
$endgroup$
– user408856
Mar 20 at 15:53
$begingroup$
There's a problem in that you still have not said carefully enough what you mean by "we identify the resulting edges". If you do not do this very carefully, there are a lot of other possible results. For example, instead of getting $S#$ (torus) you can also get $S#$(Klein bottle) (this corrects an error in the previous version of this comment). You should add to the question itself (not using a comment) exactly what you mean by gluing in a handle.
$endgroup$
– Lee Mosher
Mar 21 at 0:45