Mixing
Computing the homology groups of two tori attached in a certain way.
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Suppose $X_1$ and $X_2$ are both homeomorphic to the 2-dimensional torus $T^2$ and suppose $C_i subset X_i$ is a topological circle bounding a disc in $X_i$. Furthermore, assume a neighborhood of $C_i$ deformation retracts to $C_i$. Compute the homology of $Y$ which is constructed by gluing $X_1$ to $X_2$ along the circles $C_i$.
I want to decompose $Y=A cup B$ where $A=X_1 cup U_2$ and $B=X_2 cup U_1$ and apply Mayer-Vietors sequence to this to compute the homology groups.
With this decomposition $A cap B = U_1 cap U_2$ which is homotopy equivalent to the circle $S^1$. Also, $A,B$ are homotopy equivalent to $X_1$ and $X_2$ since $U_2$ and $U_1$ deformation retract onto the circle. My problem is that I'm not getting enough zeros in the Mayer-Vietoris sequence to get isomorphisms.
What should I do to solve this?
algebraic-topology homology-cohomology
asked yesterday
TuoTuo
1,477512
add a comment |Â
up vote
1
down vote
favorite
Suppose $X_1$ and $X_2$ are both homeomorphic to the 2-dimensional torus $T^2$ and suppose $C_i subset X_i$ is a topological circle bounding a disc in $X_i$. Furthermore, assume a neighborhood of $C_i$ deformation retracts to $C_i$. Compute the homology of $Y$ which is constructed by gluing $X_1$ to $X_2$ along the circles $C_i$.
I want to decompose $Y=A cup B$ where $A=X_1 cup U_2$ and $B=X_2 cup U_1$ and apply Mayer-Vietors sequence to this to compute the homology groups.
With this decomposition $A cap B = U_1 cap U_2$ which is homotopy equivalent to the circle $S^1$. Also, $A,B$ are homotopy equivalent to $X_1$ and $X_2$ since $U_2$ and $U_1$ deformation retract onto the circle. My problem is that I'm not getting enough zeros in the Mayer-Vietoris sequence to get isomorphisms.
What should I do to solve this?
algebraic-topology homology-cohomology
asked yesterday
TuoTuo
1,477512
Are you gluing them only along a circle, or are you gluing them along the disk they bound?
– Alfred Yerger
yesterday
@AlfredYerger Along the circle only.
– TuoTuo
yesterday
cool. So the thing that you have to remember to get this right is that you don't just need 0s in the Mayer-Vietoris sequence to know $H_i(X)$. You know some things about where the maps $i_*, j_*, k_*, l_* partial_*$ all come from because they are induced by maps of topological spaces. If you're computing purely algebraically, then you need a lot of 0s to get isomorphisms, but if you remember the topological descriptions, you can figure out what the groups are with this additional information.
– Alfred Yerger
yesterday
Also, you can work out the precise connecting homomorphism, and this will help you know the groups.
– Alfred Yerger
yesterday
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose $X_1$ and $X_2$ are both homeomorphic to the 2-dimensional torus $T^2$ and suppose $C_i subset X_i$ is a topological circle bounding a disc in $X_i$. Furthermore, assume a neighborhood of $C_i$ deformation retracts to $C_i$. Compute the homology of $Y$ which is constructed by gluing $X_1$ to $X_2$ along the circles $C_i$.
I want to decompose $Y=A cup B$ where $A=X_1 cup U_2$ and $B=X_2 cup U_1$ and apply Mayer-Vietors sequence to this to compute the homology groups.
With this decomposition $A cap B = U_1 cap U_2$ which is homotopy equivalent to the circle $S^1$. Also, $A,B$ are homotopy equivalent to $X_1$ and $X_2$ since $U_2$ and $U_1$ deformation retract onto the circle. My problem is that I'm not getting enough zeros in the Mayer-Vietoris sequence to get isomorphisms.
What should I do to solve this?
algebraic-topology homology-cohomology
asked yesterday
TuoTuo
1,477512
Suppose $X_1$ and $X_2$ are both homeomorphic to the 2-dimensional torus $T^2$ and suppose $C_i subset X_i$ is a topological circle bounding a disc in $X_i$. Furthermore, assume a neighborhood of $C_i$ deformation retracts to $C_i$. Compute the homology of $Y$ which is constructed by gluing $X_1$ to $X_2$ along the circles $C_i$.
I want to decompose $Y=A cup B$ where $A=X_1 cup U_2$ and $B=X_2 cup U_1$ and apply Mayer-Vietors sequence to this to compute the homology groups.
With this decomposition $A cap B = U_1 cap U_2$ which is homotopy equivalent to the circle $S^1$. Also, $A,B$ are homotopy equivalent to $X_1$ and $X_2$ since $U_2$ and $U_1$ deformation retract onto the circle. My problem is that I'm not getting enough zeros in the Mayer-Vietoris sequence to get isomorphisms.
What should I do to solve this?
algebraic-topology homology-cohomology
asked yesterday
TuoTuo
1,477512
asked yesterday
TuoTuo
1,477512
asked yesterday
TuoTuo
1,477512
asked yesterday
TuoTuo
1,477512
1,477512
Are you gluing them only along a circle, or are you gluing them along the disk they bound?
– Alfred Yerger
yesterday
@AlfredYerger Along the circle only.
– TuoTuo
yesterday
cool. So the thing that you have to remember to get this right is that you don't just need 0s in the Mayer-Vietoris sequence to know $H_i(X)$. You know some things about where the maps $i_*, j_*, k_*, l_* partial_*$ all come from because they are induced by maps of topological spaces. If you're computing purely algebraically, then you need a lot of 0s to get isomorphisms, but if you remember the topological descriptions, you can figure out what the groups are with this additional information.
– Alfred Yerger
yesterday
Also, you can work out the precise connecting homomorphism, and this will help you know the groups.
– Alfred Yerger
yesterday
add a comment |Â
Are you gluing them only along a circle, or are you gluing them along the disk they bound?
– Alfred Yerger
yesterday
@AlfredYerger Along the circle only.
– TuoTuo
yesterday
cool. So the thing that you have to remember to get this right is that you don't just need 0s in the Mayer-Vietoris sequence to know $H_i(X)$. You know some things about where the maps $i_*, j_*, k_*, l_* partial_*$ all come from because they are induced by maps of topological spaces. If you're computing purely algebraically, then you need a lot of 0s to get isomorphisms, but if you remember the topological descriptions, you can figure out what the groups are with this additional information.
– Alfred Yerger
yesterday
Also, you can work out the precise connecting homomorphism, and this will help you know the groups.
– Alfred Yerger
yesterday
Are you gluing them only along a circle, or are you gluing them along the disk they bound?
– Alfred Yerger
yesterday
Are you gluing them only along a circle, or are you gluing them along the disk they bound?
– Alfred Yerger
yesterday
@AlfredYerger Along the circle only.
– TuoTuo
yesterday
@AlfredYerger Along the circle only.
– TuoTuo
yesterday
cool. So the thing that you have to remember to get this right is that you don't just need 0s in the Mayer-Vietoris sequence to know $H_i(X)$. You know some things about where the maps $i_*, j_*, k_*, l_* partial_*$ all come from because they are induced by maps of topological spaces. If you're computing purely algebraically, then you need a lot of 0s to get isomorphisms, but if you remember the topological descriptions, you can figure out what the groups are with this additional information.
– Alfred Yerger
yesterday
cool. So the thing that you have to remember to get this right is that you don't just need 0s in the Mayer-Vietoris sequence to know $H_i(X)$. You know some things about where the maps $i_*, j_*, k_*, l_* partial_*$ all come from because they are induced by maps of topological spaces. If you're computing purely algebraically, then you need a lot of 0s to get isomorphisms, but if you remember the topological descriptions, you can figure out what the groups are with this additional information.
– Alfred Yerger
yesterday
Also, you can work out the precise connecting homomorphism, and this will help you know the groups.
– Alfred Yerger
yesterday
Also, you can work out the precise connecting homomorphism, and this will help you know the groups.
– Alfred Yerger
yesterday
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
Here is my attempt, where everything seems reasonable and resists my sanity checks, but I'm not an algebraic topologist so please check over my work!
The main idea is what I wrote in the comment - we figure out what the maps are in the Mayer-Vietoris sequence by looking at where they come from topologically, and use the fact that a long exact sequence can be split up into short exact sequences, and the fact that $mathbbZ$ is projective in order to stitch together the homology from the image and kernel. This wouldn't work if all the groups in sight for $A, B, U_1, U_2$ weren't all just $mathbbZ$ and powers thereof. But since they are, we get some extra computing power from this too, so we'll use your decomposition, since it's a nice one for calculating with.
So let's start with the sequence.
$$0 to H_2(A cap B) to H_2(A) oplus H_2(B) to H_2(X) to H_1(A cap B) to H_1(A) oplus H_1(B) to H_1(X) to ...$$
The tail of the sequence is kind of trivial. We know that the zeroth homology is just the free group on the connected components. All our spaces are connected, so we know that the tail of the sequence is
$$ ...to mathbbZ to mathbbZ oplus mathbbZ to mathbbZ to 0 $$
But just to sanity check, let's take a quick look at the maps.
Thus, the map taking $mathbbZ to mathbbZ oplus mathbbZ$ takes the connected component corresponding to the the cylinder and includes it into each. We can choose orientations so that this is the map $1 mapsto (1,1)$. If we do this, it follows that the map taking $mathbbZ oplus mathbbZ to mathbbZ$ must be $(n,m) mapsto n - m$, which is reasonable, since this map is $k_* - l_*$, and both $k$ and $l$ ought to be the maps taking their respective generator to itself, as both tori are connected and the whole space is connected. It is easy to see this is exact.
So now let's do this with $H_2$ since it is the next easiest, in my opinion.
You correctly observe that the intersection is homotopy equivalent to a circle. This makes $H_2(A cap B) = 0$. Also, $H_2(A) = H_2(B) = mathbbZ$ as tori are orientable surfaces, and it is a standard fact that such surfaces have top homology $mathbbZ$. Note that our space is not a topological surface.
Now we know by injectivity (by exactness) that $H_2(X)$ must contain $mathbbZ oplus mathbbZ$. We know the homology groups of the circle and tori, so we are looking at the section of the sequence that looks like
$$ ... to mathbbZ oplus mathbbZ to H_2(X) to mathbbZ to mathbbZ^4 to ... $$
and we want to what this last map is. It is the sum of two maps, each $mathbbZ to mathbbZ oplus mathbbZ$, taking the homology class of the cylinder into the torus. But our decomposition shows that this disk which it bounds is a boundary, and so this map is the $0$ map, and so this plus the algebraic observations at the beginning imply $H_2(X) = mathbbZ^3$.
This is a reasonable answer because thinking of this space as a CW complex, each torus has a contribution of a $mathbbZ$, and the gluing along a cylinder creates a sort of sphere enclosed inside the cylinder and the disks of the tori.
Now let's see about $H_1$. We expect that attaching the cylinder doesn't change anything because we already saw that the inclusions were the $0$ map, so there shouldn't be a homology class corresponding to the cylinder. To see this, observe that since $(i_*, j_*)$ is the $0$ map, the map $mathbbZ^4 to H_1(X)$ must be injective. Thus it's image is $mathbbZ^4$, which means the kernel of $partial_*$ is $mathbbZ^4$. But we already saw that the map $mathbbZ to mathbbZ oplus mathbbZ$ on $0$th homology was injective, and this means $partial_*$ has image $0$, since the next map has kernel $0$. Now we know that $partial_*$ is the $0$ map, and we know it's kernel, so we know that $H_1(X) = mathbbZ^4$ as well.
And we're done!
answered yesterday
Alfred Yerger
9,2761743
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Here is my attempt, where everything seems reasonable and resists my sanity checks, but I'm not an algebraic topologist so please check over my work!
The main idea is what I wrote in the comment - we figure out what the maps are in the Mayer-Vietoris sequence by looking at where they come from topologically, and use the fact that a long exact sequence can be split up into short exact sequences, and the fact that $mathbbZ$ is projective in order to stitch together the homology from the image and kernel. This wouldn't work if all the groups in sight for $A, B, U_1, U_2$ weren't all just $mathbbZ$ and powers thereof. But since they are, we get some extra computing power from this too, so we'll use your decomposition, since it's a nice one for calculating with.
So let's start with the sequence.
$$0 to H_2(A cap B) to H_2(A) oplus H_2(B) to H_2(X) to H_1(A cap B) to H_1(A) oplus H_1(B) to H_1(X) to ...$$
The tail of the sequence is kind of trivial. We know that the zeroth homology is just the free group on the connected components. All our spaces are connected, so we know that the tail of the sequence is
$$ ...to mathbbZ to mathbbZ oplus mathbbZ to mathbbZ to 0 $$
But just to sanity check, let's take a quick look at the maps.
Thus, the map taking $mathbbZ to mathbbZ oplus mathbbZ$ takes the connected component corresponding to the the cylinder and includes it into each. We can choose orientations so that this is the map $1 mapsto (1,1)$. If we do this, it follows that the map taking $mathbbZ oplus mathbbZ to mathbbZ$ must be $(n,m) mapsto n - m$, which is reasonable, since this map is $k_* - l_*$, and both $k$ and $l$ ought to be the maps taking their respective generator to itself, as both tori are connected and the whole space is connected. It is easy to see this is exact.
So now let's do this with $H_2$ since it is the next easiest, in my opinion.
You correctly observe that the intersection is homotopy equivalent to a circle. This makes $H_2(A cap B) = 0$. Also, $H_2(A) = H_2(B) = mathbbZ$ as tori are orientable surfaces, and it is a standard fact that such surfaces have top homology $mathbbZ$. Note that our space is not a topological surface.
Now we know by injectivity (by exactness) that $H_2(X)$ must contain $mathbbZ oplus mathbbZ$. We know the homology groups of the circle and tori, so we are looking at the section of the sequence that looks like
$$ ... to mathbbZ oplus mathbbZ to H_2(X) to mathbbZ to mathbbZ^4 to ... $$
and we want to what this last map is. It is the sum of two maps, each $mathbbZ to mathbbZ oplus mathbbZ$, taking the homology class of the cylinder into the torus. But our decomposition shows that this disk which it bounds is a boundary, and so this map is the $0$ map, and so this plus the algebraic observations at the beginning imply $H_2(X) = mathbbZ^3$.
This is a reasonable answer because thinking of this space as a CW complex, each torus has a contribution of a $mathbbZ$, and the gluing along a cylinder creates a sort of sphere enclosed inside the cylinder and the disks of the tori.
Now let's see about $H_1$. We expect that attaching the cylinder doesn't change anything because we already saw that the inclusions were the $0$ map, so there shouldn't be a homology class corresponding to the cylinder. To see this, observe that since $(i_*, j_*)$ is the $0$ map, the map $mathbbZ^4 to H_1(X)$ must be injective. Thus it's image is $mathbbZ^4$, which means the kernel of $partial_*$ is $mathbbZ^4$. But we already saw that the map $mathbbZ to mathbbZ oplus mathbbZ$ on $0$th homology was injective, and this means $partial_*$ has image $0$, since the next map has kernel $0$. Now we know that $partial_*$ is the $0$ map, and we know it's kernel, so we know that $H_1(X) = mathbbZ^4$ as well.
And we're done!
answered yesterday
Alfred Yerger
9,2761743
add a comment |Â
up vote
0
down vote
Here is my attempt, where everything seems reasonable and resists my sanity checks, but I'm not an algebraic topologist so please check over my work!
The main idea is what I wrote in the comment - we figure out what the maps are in the Mayer-Vietoris sequence by looking at where they come from topologically, and use the fact that a long exact sequence can be split up into short exact sequences, and the fact that $mathbbZ$ is projective in order to stitch together the homology from the image and kernel. This wouldn't work if all the groups in sight for $A, B, U_1, U_2$ weren't all just $mathbbZ$ and powers thereof. But since they are, we get some extra computing power from this too, so we'll use your decomposition, since it's a nice one for calculating with.
So let's start with the sequence.
$$0 to H_2(A cap B) to H_2(A) oplus H_2(B) to H_2(X) to H_1(A cap B) to H_1(A) oplus H_1(B) to H_1(X) to ...$$
The tail of the sequence is kind of trivial. We know that the zeroth homology is just the free group on the connected components. All our spaces are connected, so we know that the tail of the sequence is
$$ ...to mathbbZ to mathbbZ oplus mathbbZ to mathbbZ to 0 $$
But just to sanity check, let's take a quick look at the maps.
Thus, the map taking $mathbbZ to mathbbZ oplus mathbbZ$ takes the connected component corresponding to the the cylinder and includes it into each. We can choose orientations so that this is the map $1 mapsto (1,1)$. If we do this, it follows that the map taking $mathbbZ oplus mathbbZ to mathbbZ$ must be $(n,m) mapsto n - m$, which is reasonable, since this map is $k_* - l_*$, and both $k$ and $l$ ought to be the maps taking their respective generator to itself, as both tori are connected and the whole space is connected. It is easy to see this is exact.
So now let's do this with $H_2$ since it is the next easiest, in my opinion.
You correctly observe that the intersection is homotopy equivalent to a circle. This makes $H_2(A cap B) = 0$. Also, $H_2(A) = H_2(B) = mathbbZ$ as tori are orientable surfaces, and it is a standard fact that such surfaces have top homology $mathbbZ$. Note that our space is not a topological surface.
Now we know by injectivity (by exactness) that $H_2(X)$ must contain $mathbbZ oplus mathbbZ$. We know the homology groups of the circle and tori, so we are looking at the section of the sequence that looks like
$$ ... to mathbbZ oplus mathbbZ to H_2(X) to mathbbZ to mathbbZ^4 to ... $$
and we want to what this last map is. It is the sum of two maps, each $mathbbZ to mathbbZ oplus mathbbZ$, taking the homology class of the cylinder into the torus. But our decomposition shows that this disk which it bounds is a boundary, and so this map is the $0$ map, and so this plus the algebraic observations at the beginning imply $H_2(X) = mathbbZ^3$.
This is a reasonable answer because thinking of this space as a CW complex, each torus has a contribution of a $mathbbZ$, and the gluing along a cylinder creates a sort of sphere enclosed inside the cylinder and the disks of the tori.
Now let's see about $H_1$. We expect that attaching the cylinder doesn't change anything because we already saw that the inclusions were the $0$ map, so there shouldn't be a homology class corresponding to the cylinder. To see this, observe that since $(i_*, j_*)$ is the $0$ map, the map $mathbbZ^4 to H_1(X)$ must be injective. Thus it's image is $mathbbZ^4$, which means the kernel of $partial_*$ is $mathbbZ^4$. But we already saw that the map $mathbbZ to mathbbZ oplus mathbbZ$ on $0$th homology was injective, and this means $partial_*$ has image $0$, since the next map has kernel $0$. Now we know that $partial_*$ is the $0$ map, and we know it's kernel, so we know that $H_1(X) = mathbbZ^4$ as well.
And we're done!
answered yesterday
Alfred Yerger
9,2761743
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Here is my attempt, where everything seems reasonable and resists my sanity checks, but I'm not an algebraic topologist so please check over my work!
The main idea is what I wrote in the comment - we figure out what the maps are in the Mayer-Vietoris sequence by looking at where they come from topologically, and use the fact that a long exact sequence can be split up into short exact sequences, and the fact that $mathbbZ$ is projective in order to stitch together the homology from the image and kernel. This wouldn't work if all the groups in sight for $A, B, U_1, U_2$ weren't all just $mathbbZ$ and powers thereof. But since they are, we get some extra computing power from this too, so we'll use your decomposition, since it's a nice one for calculating with.
So let's start with the sequence.
$$0 to H_2(A cap B) to H_2(A) oplus H_2(B) to H_2(X) to H_1(A cap B) to H_1(A) oplus H_1(B) to H_1(X) to ...$$
The tail of the sequence is kind of trivial. We know that the zeroth homology is just the free group on the connected components. All our spaces are connected, so we know that the tail of the sequence is
$$ ...to mathbbZ to mathbbZ oplus mathbbZ to mathbbZ to 0 $$
But just to sanity check, let's take a quick look at the maps.
Thus, the map taking $mathbbZ to mathbbZ oplus mathbbZ$ takes the connected component corresponding to the the cylinder and includes it into each. We can choose orientations so that this is the map $1 mapsto (1,1)$. If we do this, it follows that the map taking $mathbbZ oplus mathbbZ to mathbbZ$ must be $(n,m) mapsto n - m$, which is reasonable, since this map is $k_* - l_*$, and both $k$ and $l$ ought to be the maps taking their respective generator to itself, as both tori are connected and the whole space is connected. It is easy to see this is exact.
So now let's do this with $H_2$ since it is the next easiest, in my opinion.
You correctly observe that the intersection is homotopy equivalent to a circle. This makes $H_2(A cap B) = 0$. Also, $H_2(A) = H_2(B) = mathbbZ$ as tori are orientable surfaces, and it is a standard fact that such surfaces have top homology $mathbbZ$. Note that our space is not a topological surface.
Now we know by injectivity (by exactness) that $H_2(X)$ must contain $mathbbZ oplus mathbbZ$. We know the homology groups of the circle and tori, so we are looking at the section of the sequence that looks like
$$ ... to mathbbZ oplus mathbbZ to H_2(X) to mathbbZ to mathbbZ^4 to ... $$
and we want to what this last map is. It is the sum of two maps, each $mathbbZ to mathbbZ oplus mathbbZ$, taking the homology class of the cylinder into the torus. But our decomposition shows that this disk which it bounds is a boundary, and so this map is the $0$ map, and so this plus the algebraic observations at the beginning imply $H_2(X) = mathbbZ^3$.
This is a reasonable answer because thinking of this space as a CW complex, each torus has a contribution of a $mathbbZ$, and the gluing along a cylinder creates a sort of sphere enclosed inside the cylinder and the disks of the tori.
Now let's see about $H_1$. We expect that attaching the cylinder doesn't change anything because we already saw that the inclusions were the $0$ map, so there shouldn't be a homology class corresponding to the cylinder. To see this, observe that since $(i_*, j_*)$ is the $0$ map, the map $mathbbZ^4 to H_1(X)$ must be injective. Thus it's image is $mathbbZ^4$, which means the kernel of $partial_*$ is $mathbbZ^4$. But we already saw that the map $mathbbZ to mathbbZ oplus mathbbZ$ on $0$th homology was injective, and this means $partial_*$ has image $0$, since the next map has kernel $0$. Now we know that $partial_*$ is the $0$ map, and we know it's kernel, so we know that $H_1(X) = mathbbZ^4$ as well.
And we're done!
answered yesterday
Alfred Yerger
9,2761743
Here is my attempt, where everything seems reasonable and resists my sanity checks, but I'm not an algebraic topologist so please check over my work!
The main idea is what I wrote in the comment - we figure out what the maps are in the Mayer-Vietoris sequence by looking at where they come from topologically, and use the fact that a long exact sequence can be split up into short exact sequences, and the fact that $mathbbZ$ is projective in order to stitch together the homology from the image and kernel. This wouldn't work if all the groups in sight for $A, B, U_1, U_2$ weren't all just $mathbbZ$ and powers thereof. But since they are, we get some extra computing power from this too, so we'll use your decomposition, since it's a nice one for calculating with.
So let's start with the sequence.
$$0 to H_2(A cap B) to H_2(A) oplus H_2(B) to H_2(X) to H_1(A cap B) to H_1(A) oplus H_1(B) to H_1(X) to ...$$
The tail of the sequence is kind of trivial. We know that the zeroth homology is just the free group on the connected components. All our spaces are connected, so we know that the tail of the sequence is
$$ ...to mathbbZ to mathbbZ oplus mathbbZ to mathbbZ to 0 $$
But just to sanity check, let's take a quick look at the maps.
Thus, the map taking $mathbbZ to mathbbZ oplus mathbbZ$ takes the connected component corresponding to the the cylinder and includes it into each. We can choose orientations so that this is the map $1 mapsto (1,1)$. If we do this, it follows that the map taking $mathbbZ oplus mathbbZ to mathbbZ$ must be $(n,m) mapsto n - m$, which is reasonable, since this map is $k_* - l_*$, and both $k$ and $l$ ought to be the maps taking their respective generator to itself, as both tori are connected and the whole space is connected. It is easy to see this is exact.
So now let's do this with $H_2$ since it is the next easiest, in my opinion.
You correctly observe that the intersection is homotopy equivalent to a circle. This makes $H_2(A cap B) = 0$. Also, $H_2(A) = H_2(B) = mathbbZ$ as tori are orientable surfaces, and it is a standard fact that such surfaces have top homology $mathbbZ$. Note that our space is not a topological surface.
Now we know by injectivity (by exactness) that $H_2(X)$ must contain $mathbbZ oplus mathbbZ$. We know the homology groups of the circle and tori, so we are looking at the section of the sequence that looks like
$$ ... to mathbbZ oplus mathbbZ to H_2(X) to mathbbZ to mathbbZ^4 to ... $$
and we want to what this last map is. It is the sum of two maps, each $mathbbZ to mathbbZ oplus mathbbZ$, taking the homology class of the cylinder into the torus. But our decomposition shows that this disk which it bounds is a boundary, and so this map is the $0$ map, and so this plus the algebraic observations at the beginning imply $H_2(X) = mathbbZ^3$.
This is a reasonable answer because thinking of this space as a CW complex, each torus has a contribution of a $mathbbZ$, and the gluing along a cylinder creates a sort of sphere enclosed inside the cylinder and the disks of the tori.
Now let's see about $H_1$. We expect that attaching the cylinder doesn't change anything because we already saw that the inclusions were the $0$ map, so there shouldn't be a homology class corresponding to the cylinder. To see this, observe that since $(i_*, j_*)$ is the $0$ map, the map $mathbbZ^4 to H_1(X)$ must be injective. Thus it's image is $mathbbZ^4$, which means the kernel of $partial_*$ is $mathbbZ^4$. But we already saw that the map $mathbbZ to mathbbZ oplus mathbbZ$ on $0$th homology was injective, and this means $partial_*$ has image $0$, since the next map has kernel $0$. Now we know that $partial_*$ is the $0$ map, and we know it's kernel, so we know that $H_1(X) = mathbbZ^4$ as well.
And we're done!
answered yesterday
Alfred Yerger
9,2761743
answered yesterday
Alfred Yerger
9,2761743
answered yesterday
Alfred Yerger
9,2761743
answered yesterday
Alfred Yerger
9,2761743
9,2761743
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2872261%2fcomputing-the-homology-groups-of-two-tori-attached-in-a-certain-way%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Are you gluing them only along a circle, or are you gluing them along the disk they bound?
– Alfred Yerger
yesterday
@AlfredYerger Along the circle only.
– TuoTuo
yesterday
cool. So the thing that you have to remember to get this right is that you don't just need 0s in the Mayer-Vietoris sequence to know $H_i(X)$. You know some things about where the maps $i_*, j_*, k_*, l_* partial_*$ all come from because they are induced by maps of topological spaces. If you're computing purely algebraically, then you need a lot of 0s to get isomorphisms, but if you remember the topological descriptions, you can figure out what the groups are with this additional information.
– Alfred Yerger
yesterday
Also, you can work out the precise connecting homomorphism, and this will help you know the groups.
– Alfred Yerger
yesterday