A
Subdivide the surface into triangles. Find maps from the standard 2-simplex onto
the triangles, in such a way that
when you take the boundary of the chain formed by the
sum of these maps, you get a 1-chain lying in the boundary
of the surface. In fact the exercise asks slightly more
- we want this 1-chain to have a consistent orientation,
in the
sense that it is a generator of the first homology of the boundary. If you
use the model of the torus as quotient of a
square, you can specify maps from the standard
2-simplex to the torus simply by listing the vertices. This gives six
possibilities for
each triangle. Trial and error is therefore a possibility.
As you proceed you will find yourself
developing a more scientific approach, observing how to order the vertices so
that internal edges cancel out, as we
did when we computed the delta complex structure and
simplicial homology of the torus and projective plane.
A If the polynomial is f(z)=z^n, then the local degree of f^ at infinity
is not hard to see, e.g. by restricting to a cicle S_R centred
at 0 with
radius R. Now for a general degree n polynomial f, if R is big enough the
restriction of f to S_R is homotopic to the
restriction of z^n to S_R.
Moral: always try the easiest examples first.
NN wrote:
I have a question about the first homework sheet. I got it wrong, but I
don't understand why it is wrong.
In question 4.ii
we have to show that the short exact sequence of
abelian groups
0->A->B->Z^n->0
implies that
B = Z^n + A. The way I did it is the following, we know that A->B is
injective and we know that B->Z^n
is surjective. Can we then use the 1st
isomorphism theorem saying that B/A = Z^n (because doesn't the statement
then just follow)? Probably not but I can't see why not.
Dear NN
You are right that the first isomorphism theorem says that B/A=Z^n.
But this does not imply that B=A+Z^n. Let's take
an example:
Consider the short exact sequence
0 --> Z --> Z --> Z/2Z -->0
where the map Z --> Z is multiplication by 2, and the map Z--> Z/2Z
is just the projection to the quotient. The sequence is
clearly exact, but we do not have
Z = Z + Z/2Z.
Every short exact sequence
(*) 0 --> A --> B --> C --> 0
leads to the correct statement "B/A=C" (or more precisely
"B/A is isomorphic to C") but only in some cases can one make
the extra step to
"B=A+C".
When this extra step is correct, we say the short exact sequence
"splits". The aim of the exercise is to show that any short
exact sequence (*) in which C=Z^n, does indeed split.
Brief sketch of the argument:
if (*) is exact and C=Z^n, let
e_1, ..., e_n be generators of Z^n. Because B --> C is surjective, there
exist b_1,...,b_n in B
mapping to e_1, ..., e_n. Define a map C-->B by sending
e_i to b_i for i=1,...,n, and then "extending by linearity": every element
c of C is a unique linear combination c= p_1e_1+...+p_ne_n where the p_i are integers. We send c to p_1b_1+...+p_nb_n.
(The crucial word in the last paragraph is the word "unique".)
You should check that we have defined a homomorphism C-->B which is,
moreover, a right inverse to the map B-->C. Let us
call these two maps f and j respectively. So we have jf=identity on C.
Next, show that for any b in B, we have
b-fj(b) is in kernel of j.
By exactness of (*) this means that b-fj(b) is in the image of i:A--> B.
Define a map g:B -->A by sending b to the unique element a of A such that
i(a)=b-fj(b). Then we have a map B to A+C sending b
to (g(b), j(b)). This map is an isomorphism.
In question 4 of homework 2, I tried to calculate the reduced homology
groups of X using the Mayer-Vietoris l.e.s. and
the following data
[in which H_k denotes reduced homology because D.M. does not know how to write
a tilde in html]
H_k(R)=Z^g if k=1 and 0 otherwise
H_k(M_g)= Z if k=2, Z^2g if k=1 and 0 otherwise
From this I worked out that H_3(X)=Z and H_0(X)=0. But how can I compute H_1 and H_2?
Answer
Let R_1 and R_2 denote the two copies of the compact region enclosed by the
surface M_g. The portion of the
Mayer Vietoris sequence which requires thought is
(*) 0--> H_2(X)-->H_1(R_1\cap R_2)\to H_1(R_1)+H_1(R_2)--> H_1(X)-->0
Now R_1\cap R_2 is equal to M_g, whose homology we know.
What about H_1(R_1) and H_1(R_2)? These can be obtained
from
H_1(M_g) by "killing
half of the homology classes". If you draw the standard picture of M_g, lying
horizontally on a table,
there is a basis for H_1(M_g) consisting of
cycles of two types: horizontal ones (lying in a horizontal plane parallel to
the table
top) and vertical ones (each one lying in a vertical plane
, not necessarily the same plane each time). See for example the
picture of M_2 on page 132 of Hatcher. It shows two vertical
cycles and no horizontal cycles. When we "fill in" M_g to get the
compact region in R^3 that it bounds, R_1,
the
homology classes of the vertical cycles die (become 0), while the
homology
classes of the horizontal cycles survive.
There are g cycles of each
type in the basis of H_1(M_g). So
H_1(R_1)=Z^g.
Draw the picture! On the other hand,
when we fill in the "outside" of M_g to get R_2, it is
the horizontal cycles which die.
This is less obvious.
Putting this information into
the exact sequence (*) above,
it is possible to compute H_2(X) and H_1(X).
The relative homology
groups can also be calculated using the same approach.
On question 3c) I think I've deduced that the induced map must be
the identity on H_3(X), H_0(X) and H_2(X) (which is also the
zero map since H_2(X)=0.) For the induced map on H_1(x)=Z/2Z +
Z/2Z there are only two options: either it is the identity or
it sends each generator to the other. Is there a
quick way to check which it is or do you have to do a lot of algebra
to check?
Answer
The map f sends $X^k$ to itself for each $k$, and therefore
induces a homomorphism of the cellular chain complex; i.e.,
induces
homomorphisms
H_k(X^k,X^{k-1}) --> H_k(X^k, X^{k-1})
for
k=0,1,2,3, which commute with the differentials. If you look
carefully at what these do to the generators of each of these groups,
you can figure out what $f_*$ does
to the homology of X.
I 'm confused about computing the homology group of the Mobius band
relative to its bounding ciricle (i.e. H_1(M,S^1)). Is there a way to
do it through the delta complex structure we give to the Mobius band?
Throughout Hatcher it is mentioned that "the bounding circle of a
Mobius band wraps twice around its core circle". I understand that but
how do I prove it? Because if I can say that f:S^1----> M is a degree
2 map, then the isomorphism H_1(S^1)--->H_1(M) is given by f(a)=2a, so I
can use the long exact sequence of the pair (M,S^1) to compute
the relative homology.
Answer
Represent the Mobius band as the rectangle [0,2]x[-1,1]
subject to the identification [0,t]~[2,-t] for t in [-1,1].
The core circle C_c is [0,2]x{0},
subject to the identification (0,0)~(2,0).
The bounding circle C_b is
[0,2]x{-1,1} subject to the identification (0,-1)~(2,1) and (0,1)~(2,-1).
Draw a picture!
The map from bounding circle to core circle sends (x,y) to (x,0).
This map is obviously 2 to 1. It remains to compute its degree. Both core
circle and bounding circle are homeomorphic to S^1, but there is no canonical
homeomorphism in either case. Thus the degree is defined only up to sign.
To compute it, use the local degree as described in Hatcher on page 136.
Choose a point in the core circle - for example (1,0) - and compute local degree
at each of its preimages. The preimages are the points (1,1) and (1,-1).
Depending on how we orient the bounding and core circles (i.e. which
isomorphism H_1(C_c)-->Z and H_1(C_b)-->Z we choose), both local contributions
are +1, or both are -1. So the degree is 2 or -2.
I'm explicitly calculating the image of the boundary maps in homology.
In the revision lecture we did this using elementary row and column
operations on the matrix. Is this a sufficient way of doing this or
is this just intuitive and we should explicitly construct an isomorphism
from the group generated by the image of the generators, to the group
we get from row and column operations?
Answer
Let
...--> C_{n+1} --> C_n --> C_{n-1} --> ...
be a complex of finitely generated free abelian groups (i.e. each is isomorphic
to Z^k for some finite k). Every subgroup of a free abelian group is
free abelian. So the kernel of d:C_n --> C_{n-1} is a free abelian group.
Call it K. Choose a basis, k_1, ...,k_m for K. Let b_1,...,b_r be a basis
for C_{n+1}. Then d(b_i) is in K for each i, so d(b_i) is a (unique) linear
combination of the k_j. D9B_i)=a_{i1}k_1+...+a_{im}k_m. Represent
each element c_1k_1+...=c_mk_m of K as the column vector (c_1,...,c_m)^t.
K is the set of all these column vectors. d(C_{n+1}) is the subspace
generated by the column vectors (a_{i1},...,a_{im})^t for i=1,...,r. Put
these vectors together to form a matrix with m rows and r columns.
The method we used in lectures says that the quotient K/Im(d) is unchanged
(up to isomorphism) by row and column operations on this matrix.
Justification: Column operations obviously do not change the subspace
generated by the columns. So they leave the quotient group unchanged.
Row operations correspond to choosing a different basis for K. If I add,
say, row 1 to row 2, the new columns are the expressions of the d(b_i)
in the new basis k_1-k_2, k_2, k_3,...,k_m. So the quotient is the same.
If you like, a choice of basis for K is the same thing as a choice of
isomorphism Z^m --> K.