A few general remarks.
First, it is wise to look at the derivation of the simplicial homology
of the torus on Example 2.3 of Hatcher, as that gives a very concrete
example of how one can compute homology. We could do the same thing
here, but the simplices involved are a bit of a hassle to deal with
hand, although by no means impossible.
First, I will just give the answer:
The answer is:
H_0(X,B)=Z
H_1(X,B)=Z x Z x Z
H_2(X,B)=Z
H_k(X,B)=0 for k>2 .
I will give two proofs of this fact: one a concrete one based on
triangulation, and one a more abstract one based on exact sequences.
Particularly the more abstract proof may be a bit terse. Please ask
for clarifcations to clear up specific questions.
Preliminary definitions:
------------------------
Let T be a torus.
X is the connected sum of two tori, T#T, formed by cutting out a small
disk in each torus and gluing the sides together.
B is a small circle going around one of the tori.
Homology is singular homology with coefficient group Z.
First proof
-----------
This relies on a web-based simplicial homology computation program.
I will write R_n for
~
H_n
, the reduced homology functor.
We first observe that
H_n(X,B) = R_n(X/B) by proposition 2.22 of Hatcher, on page 124. Here,
(X,B) form a "good pair" as defined on page 114 following the
statement of theorem 2.13, since B is manifestly a deformation retract
of a small cylinder containing it in X.
We shall now construct a simplicial complex equal to X/B , and run
that through the homology program. This particular program requires
that its simplices be uniquely determined by its vertices, so we need
to put in some extra triangles compared to the Hatcher treatment.
First, let us triangularize the torus.
We trisect the horizontal and vertical lines in a square, and label
the intersections as follows:
1 4 7 1
2 5 8 2
3 6 9 3
1 4 7 1
Now we draw lines between horizontally and vertically adjacent nodes,
and bisect each rectangle. The triangles formed are thus:
124,245,457,578,178,128,
235,356,568,689,289,239,
136,146,469,479,137,379
Here each three digit string denotes a 2-simplex (a triangle) in the
torus.
Now let us consider the topology of Y=T/B . This represents a torus
with the wrapping circle B squashed to a point: imagine that B is a
rope, and tighten the rope until it is a point.
The resulting topology looks like two open cones - say ice-cream
cones, with their points glued together.
We now triangularize Y.
We name the vertex of Y 10 . 10 is part of the boundary of a
boundary([10,20,30,40]). Thus, to get one open cone we add faces:
[10,20,30], [10,20,40],[10,30,40]
Now we build the other open cone with faces:
[10,50,60]
[10,60,70],
[10,50,70]
Now we connect the corresponding vertices of the two triangles two
form a prism with three rectangles:
20,50,60,30
30,60,70,40
20,50,70,40
Finally, we bisect each rectangle to get the triangularization of the
prism:
[20,30,50], [30,50,60]
[30,40,60], [40,60,70]
[40,50,70], [20,40,50]
Thus, the full triangularization of T/B is:
[10,20,30], [10,20,40],[10,30,40]
[10,50,60]
[10,60,70]
[10,50,70]
[20,30,50], [30,50,60]
[30,40,60], [40,60,70]
[40,50,70], [20,40,50]
Finally, to form X/B we form the connected sum of T/B and T . We
choose a small two-simplex in T, say
[1,2,4]
and we glue that to a small triangle in T/B, say
[20,30,40]
To do this, we form the union of the faces in T/B and T, we remove
[1,2,4] and [20,30,40] from the list, and we replace respectively
20 by 1,
30 by 2
40 by 4 .
The resulting simplex is then:
[2,4,5],[4,5,7],[5,7,8],[1,7,8],[1,2,8],
[2,3,5],[3,5,6],[5,6,8],[6,8,9],[2,8,9],[2,3,9],
[1,3,6],[1,4,6],[4,6,9],[7,4,9],[3,7,9],[1,3,7],
[10,1,2],[10,1,40],[10,2,40],
[2,4,60],[2,40,60],[40,60,70],[40,4,70],[1,40,4],
[10,4,60],[10,60,70],[10,4,70]
The Homology computation server URL at:
http://aphrodite.cis.udel.edu:8080/gap/SimplicialHomologyForm.html
is a bit difficult to use. We create a URL with this string in it:
SimplicialHomology([
[2,4,5],[4,5,7],[5,7,8],[1,7,8],[1,2,8],
[2,3,5],[3,5,6],[5,6,8],[6,8,9],[2,8,9],[2,3,9],
[1,3,6],[1,4,6],[4,6,9],[7,4,9],[3,7,9],[1,3,7],
[10,1,2],[10,1,40],[10,2,40],
[2,4,60],[2,40,60],[40,60,70],[40,4,70],[1,40,4],
[10,4,60],[10,60,70],[10,4,70]
]);
say foo.gap . We then input this URL into the form at the homology
computation server. It's output looks like this:
[[0],[3],[1]]
which indicates that the reduced homology groups of (X/B) are:
R_0(X/B)=0
R_1(X/B)=Z+Z+Z [direct sum]
R_2(X/B)=Z
The result now follows.
SECOND PROOF
------------
First let us compute the homology of Y.
We see that Y is homeomorphic to a two-sphere S^2 with two points
identified. Let V be S^2 and let W={a,b} be two points on S^2.
Y=V/W
Now, (V,W) is a good pair, since W is a deformation retract of two
small disks about a and b. Therefore by theorem 2.13
we have an exact sequence:
R_2(W)->R_2(V)->R_2(Y)->R_1(W)->R_1(V)-> R_1(Y)->R_0(W)->R_0(V).
Now, R_2(W)=R_1(W)=0 since W has just 0-complexes.
R_0(W)=Z since H_0(W)=Z+Z .
We know the reduced homology of V=S^2 is 0 except at dimension 2 (this
is a standard theorem).
Therefore, the above exact sequence reduces to:
0->Z->R_2(Y)->0->0->R_1(Y)->Z->0
In any exact sequence of the form
0->G->H->0
the middle map must be an isomorphism.
Therefore, R_2(Y)=Z and R_1(Y)=Z ; we already know R_0(Y)=Z .
So we have computed the homology of Y.
Now, let X/B = Y # T = M .
Let A be a circle around the connecting bridge of M, as pictured in
the diagram in Hatcher. Note that A is homotopic to a constant map.
Now, M/A is just the wedge sum of T and Y . Therefore, the homology of
M/A is given by (following corollary 2.25)
R_0(M/A)=0
R_1(M/A)=R_1(T)+R_1(Y)=(Z+Z)+Z
R_2(M/A)=R_2(T)+R_2(Y)= Z+Z
since the reduced homology of T in dimensions 0, 1, and 2 is 0, Z^4, Z
respectively by Example 2.3 and the equivalence of singular and
simplicial homology.
Hence we have exact sequences:
R_2(A)->R_2(M)->R_2(M/A)->R_1(A)->R_1(M)->R_1(M/A)->0
The last term is 0 since the spaces are all path connected so have
trivial reduced homology.
We have R_2(A)=0 since A is S^1 .
R_2(M/A)=Z+Z.
R_1(Z)=Z since A is S^1
R_1(M/A)=Z^3
So we now have exact sequence:
0->R_2(M)->Z+Z->Z->R_1(M)->Z^3->0
The map Z->R_1(M) is induced by the inclusion map A->M. But A is
homotopic to a constant map, so that the image of this map is 0.
Hence, the map from R_1(M) to Z^3 is an isomorphism, as it is
surjective by the 0 after Z^3 and injective because its kernel is 0,
so that R_1(M)=Z^3 .
There is an injection from R_2(M) into Z+Z, so R_2(M) has no torsion
(no elements a multiple of which are zero) .
Since the map h from Z+Z to Z is surjective (as the map from Z to
R_1(M) is the 0 map), we have Z is isomorphic to (Z+Z)/ker h. This is
only possible if ker h is equal to Z, so the image of R_2(M) is equal
to Z. Since that map is an injection R_2(M) is Z.
The result now follows.
SEARCH STRATEGY
---------------
I found this far and away the trickiest of the challenges you have
set; it's solution required quite some work. Although in your note you
mentioned that it may not seem that difficult, since it is one part of
one part of an exercise, all the other parts are trivial.
[E.g., for 17(a): when one removes a point from S^2 or S^1xS^1 one
gets a disk. The homology of H_n(X,A) is virtually immediate, since
X/A is a wedge sum of tori.]
Generally in algebraic topology, one can compute homology either by
relying on powerful theorems (exactness arguments) or doing explicit
computation. I did both here, as a check.
The homology computation server was found by searching on "homology
computation". The actual server has a rather byzantine interface, but
to install it locally requires GAP (a nice package I have used before)
and both g++ and GNU Pascall (?!) . |