I am looking for a proof of the following:

 1  1      1  1  2
S xS  and S VS VS  have isomorphic homology groups in all dimensions,
but their universal covering spaces do not.

           n                                                     n+1
Note that S  is the n-dimensional sphere, that is the subset of R    
                                  2      2
homeomorphic to {(x ,...,x   ) : x +...+x   =1}.  Also,
                   1      n+1     1      n+1

AxB is the Cartesian product of A and B,

and AVB is the wedge of A and B.

If you wish you may use the following fact without proof:

    n
H (S ) = 0, if 0 < i < n or i > n
 i       Z, if i = 0 or i = n

where Z is the integers.

Also, feel free to appeal to standard facts from homology theory such as
those found in Chapter 2 of Hatcher's Algebraic Topology (a link to
this text may be found in my other question on homology theory posted on 12/6).

hi,

two facts which you may find useful:

1. According to Vick, the homology group of S^m x S^n has four free
abelian generators, of dimensions 0,m,n, and m+n. So for the torus
S^1xS^1 we have
H_0 = Z
H_1 = Z+Z
H_2 = Z
H_3 = H_4 = ... = 0

2. The universal covering space of the torus is R^2, and therefore has
homology groups H_0=Z, H_1=H_2=...=0.

Can you clarify the meaning of S^1 v S^1 v S^2 ? I can't find the
definition of wedge product in Hatcher. Vick has a definition which
confuses me a bit when it comes to taking the product of three spaces
- is it the space obtained when you attach a circle, a circle and a
2-sphere at one mutual point?

dannidin

Hi.
With the comments of dannidin, there are two points to add:
1.
Since X=S^1 v S^1 v S^2 is connected, H_0(X) = Z.
Further, the fundamental group (by (Seifert)-van Kampen's Theorem,
p.43,1.20 in Hatcher's book) is Z*Z*0 (free group product of Z with Z
with the trivial group).
Since H_1(X)=pi_1^ab=Z^2 (H_1=abelianized Pi ("small Hurewitz
theorem") for connected spaces (chapter 2.A in Hatcher)).
H_2(X) = Z (if needed by an obvious CW-construction (1 0-cell,2
1-cells,1 2-cell).
H_k(X) = 0 for k>=3 (CW dimension argument).
2.
Universal covering for X:
This can be constructed as follows: Take the Cayley-Graph of Z*Z (cf.
Hatcher p.59) and attach a S^2 at every Knot (with its "basepoint").
This is a covering space of X as only the "knots" need to be checked
and this works as on p.59).
Now, the space is clearly connected and simply connected (take a loop,
shrink it to the basepoint of each sphere it touches, this is a
homotopy, then argue as on p.59), thus it is a universal covering
space for X.
Homology: H_0=Z clear, H_1=0=pi_1^ab, H_k=0 for k>=3 (Dimension of CW
construction).
Another CW-argument (I think the CW-construction I have in mind is
obvious)
shows that H^2 contains an infinite direct Sum of Z's. (One for each
Sphere, the cellular chains (corresponding to the generator of
C_2(S^2) on each copy of S^2 in the covering space) are obviously
cycles, but they cannot be boundaries since the dimension of the
CW-complex is 2).

I guess now we know that the H^2's of the covering spaces differ, even
though we have not completely computed the second one.
I don't know whether this is all you need, so this is just a comment.

Regards

T.

Hello Dannidin and Tomvman17,

   Thank you for your comments.  It looks as if all the elements of a
solution are essentially there.  While I genuinely appreciate your
contribution, Dannidin, I suppose the question ought to go to
tomvman17 as most of what I need clarified pertains to what he wrote. 
However, if you want to tackle your fears and solve the remaining
question, I'll leave a tip on that one to acknowledge your
contribution here.
   Tomvman17, I'll give you credit for the answer, but I would be
quite grateful if you could include considerably more detail (as if
you were an undergrad writing this up to be graded).  For example, why
is the universal covering space of the torus R^2, why does the
abelianization of pi_1 equal pi_1, write out all the CW arguments,
elaborate on the proof that H_2(universal covering space for the
wedge) is non-zero, etc.  The only thing which does not need
elaboration is the computation of the homology group of the torus, as
I have found a proof of the S^m x S^n result.
   If you could do this within the next 24 hours, that would be great.
 BTW, I am reducing the asking price and will leave you the amount you
would have received in the form of a tip, as was my arrangement with
Dannidin.  Thanks.

Hi. 
I don't quite know how detailed it has to be, but here is a try:
R^2 is the universal covering space of T=S^1xS^1=R^2/Z^2 because it is
simply connected and a covering space (since T is a quotient of the
properly discontinous action of Z^2 (translation on R^2).
H_1 of X=S^1vS^1vS^2
> why does the abelianization of pi_1 equal pi_1
?. Why is H_1=pi_1^ab for connected paces? This is a theorem of
Hurewitz, best quoted (See Hatcher 2.A, it also in Bredon IV.3 (where
I learnt it from) as one of the first items on Homology).
You can also argue directly by Hatcher p. 126, Corollary 2.25 that
H_1(X)=H_1(S^1)+H_1(S^1)+H_1(S^2)=Z+Z+0.
H_k(X) = 0 for k>=3 (construct CW-Complex with 1 0-Cell (Basepoint), 2
1-Cells and 1 2-Cell (for the S^1s and S^2 respectively). Now you
don't have (k>=3)-Cells, thus H_k(X)=0 for k>=0 (If this isn't counted
as commonly known: Hatcher Chapter 2 Lemma 2.34 p. 137 - I primarily
quote Hatcher because this seems to be your preferred book).

Y = Universal covering space (Cayley-graph of Z*Z with spheres
attached).
CW-construction: Take one 0-cell for earch word in Z*Z (i.e. for each
knot in the graph). Take 1-cells for the edges of the graph. Attach a
2-cell to each knot with the attaching map constant to the knot.
This is connected, thus H_0=Z.
This is simply connected: If a loop is in any spheres, you can
homotope that section of the path to the basepoint (because S^2 is
simply connected), thus it is now a loop in the Cayley-Graph. For this
Hatcher gives the contraction on p. 59.
Thus H_1(Y)=(Hurewitz)pi_1(Y)=0
Y is a connected and simply connected covering space of X, thus a
universal covering space.
Since the CW-Dimension of Y is 2, H_k=0 for k>=3 as above.
Now H_2 (sorry to have written H^2 in my previous comment...):
If you don't like the CW-argument, consider the following segment of
long exact homology squence of the CW-pair (Y^2,Y^1) where Y^k is the
k-skeleton:

H_2(Y^1)->H_2(Y^2)->H_2(Y^2,Y^1)->H_1(Y_1)

Now H_2(Y^1)=0 (again Hatcher Ch. 2 Lemma 2.34 p. 137).
H_2(Y^2,Y^1) is free abelian with basis Z*Z (same lemma)
Note that Y^1 is the Cayley-Graph of Z*Z = Universal Covering of
S^1vS^1:
H_1(Y^1) = pi_1(Y^1)^ab (Hurewitz) = 0 because it is simply connected.
Thus 0->H_2(Y^2)->(Free abelian group with Basis Z*Z)->0, i.e. we get
an Isomorphism. (I've previously refrained from claiming Iso because I
remember a similar excercise where one had to be careful.)

Is that detailed enough? I've never been an undergrad who writes up
things to be scored, so I cannot really tell.

Regards

T.

Looks good.  If you clarify the points mentioned below and formally
post it as an answer, I'll leave you your money.  Points I would like
clarification on:

1) You compute H_2(Y) by proving H_2(Y^2) is isomorphic to the free
abelian group with basis Z*Z; why is the 2-skeleton of Y equal to Y?

2) Please explicitly write out the CW argument for computing H_2(Y)
which gives this same result.

3) Could you either write out or refer me to a proof for the homology
groups of R^2 ?

Thank you.

dubois,

regarding item 3 on your list of requests for clarification: R^2 is
contractible, so it has the homotopy type of a point, and therefore
the same homology groups as a point: H_0=Z (because of one connected
component) and all H_n, n>0 are 0 (on a one point space there is only
one singular simplex of each dimension so it's very easy to compute).

by the way i'm not trying to stake a claim on this question, tomvman17
earned it fair and square... i'm glad to see you found an answer to
all of your AT questions.

cheers,
dannidin

This leaves 1 and 2:
ad 1
Since the k-Skeleton of a CW-Complex is the subcomplex of the cells
with Dimension<=k, for k=dimension(=highest dimension of cell=here 2)
you end up with the CW-Complex.
ad 2
I will remark that the exact sequence argument seems quicker and
nicer, but here you go:
The attaching map S^2=boundary(D^2) of each 2-cell is constant to a
0-cell (thus having "degree 0" in the sense of the cellular boundary
formula p.140 of Hatcher). Therefore the induced differential C_2->C_1
in the cellular chain complex is 0. Since C_3=0 (there are no 3-cells
in the CW-complex), we have that the cellular homology
H^{CW}_2(Y)=C_2(Y)=(free abelian group generated by Z*Z), and, of
course H^{CW}=H (p.139 Thm. 2.35).
.

Tomvman17, everything looks quite good.  Click answer and I'll give
you your rating and tip.  Dannidin, thank you for all your help.  I
really appreciate it, and I'll acknowledge it if you answer future
questions of mine.

I am grateful to all the math researchers for solving all of my
questions.

Hm. Sorry. Just found out, that there is no answer button.
Regards
T.

Tomvman17, when you say that there is no answer button, do you mean
that there is a problem with Google's software, or that you are not
actually a researcher?

Yes, upon closer inspection, I'm not a researcher.
Either I'm a scam artist or I reaffirm anyone in the prejudice that
mathematicians get carried away with mathematics to not care about the
simple things.
I only registered at google answers when it was brand new and always
had thought about it more as a peer service...
Anyhow. I hope the answers were still of value to you.
Regards
T.

Please pardon my last comment; I certainly was not trying to accuse
you of anything, rather, I was slightly puzzled.  I am sorry that the
system is constructed such that I cannot compensate you in any way. 
However, your answers were of the greatest use to me and I am
extremely grateful for your time and generosity.  I owe you one!