Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What's the difference between $C_n(A + B)$ (that is, $C_n(A)+C_n(B)$ in $C_n(X)$) and $C_n(A) \oplus C_n(B)$ where $A$ and $B$ are subspaces of a topological space $X$? They're the same sets, right?

Thanks for your help.

share|improve this question
    
I have taken this notation from Hatcher p. 149, I think $A + B$ means $a + b$ for chains $a \in A$ and $b \in B$ where $a = \sum_i n_i \sigma_i$ and $b = \sum_i m_i \tau_i$ –  Matt N. Aug 9 '11 at 11:41
    
Actually, I think elements in $C(A) \oplus C(B)$ look like $(a,b)$ and the corresponding element in $C(A+B)$ looks like $a+b$... but they represent the same element –  Matt N. Aug 9 '11 at 11:46
4  
@Grigory: This is not really Matt's fault, I think. Hatcher introduces the notation $C(A+B)$ for those elements of $C(X)$ that are sums of elements of $C(A)$ and $C(B)$ (and Mayer-Vietoris is the long exact sequence associated to the short exact sequence $0 \to C(A \cap B) \to C(A) \oplus C(B) \to C(A + B) \to 0$ of course, where the first map is the signed inclusion and the second is the sum). On the other hand, this short exact sequence clearly says what the difference between $C(A) \oplus C(B)$ and $C(A + B)$ is... –  t.b. Aug 9 '11 at 12:04
2  
That's an awful notation (IMHO) –  Andrea Mori Aug 9 '11 at 12:35
1  
@Grigory: Yes exactly. I did repeat what Hatcher says (in order to avoid confusing Matt even more) and written this way it really is confusing. I do see your point on "research effort" and I didn't upvote this question, but I'm generally quite happy if the OP honestly tries to learn something on his own and tries to get help here. That's what Matt obviously does in view of the questions he asks. In those cases I try to be a bit more lenient and refrain from downvoting in order not to discourage them. –  t.b. Aug 9 '11 at 12:57
show 5 more comments

1 Answer 1

up vote 10 down vote accepted

The point is that $C(A + B)$ is the subgroup of $C(A \cup B)$ generated by $C(A)$ and $C(B)$ while $C(A) \oplus C(B)$ is the abelian group consisting of pairs of elements $(c_1,c_2)$ with $c_1 \in C(A)$ and $c_2 \in C(B)$. The elements that $C(A)$ and $C(B)$ have in common (that is: exactly the elements of $C(A \cap B)$) occur twice in the direct sum $C(A) \oplus C(B)$: once as $(c,0)$ and once as $(0,c)$. On the other hand $C(A + B)$ is by definition the subgroup of all elements $c \in C(A \cup B)$ that can be written as the sum $c = c_1 + c_2$ with $c_1 \in C(A)$ and $c_2 \in C(B)$. In particular $C(A) \oplus C(B)$ and $C(A + B)$ are different (unless $A \cap B = \emptyset$).


Since I told you many times already that you should try and draw some pictures, here's how I would "see" Mayer-Vietoris:

Since the boundary map of $C(A \cup B)$ restricts to $C(A + B)$ (why?), we see that $C(A + B)$ is a chain complex.

Note that it is not all of $C(A \cup B)$ since a simplex $\sigma$ in $A \cup B$ can look as follows:

simplex

However, after (possibly a few) barycenric subdivisions, each such simplex will be decomposed into (more formally: chain equivalent to a sum of) simplices that lie entirely inside either $A$ or $B$. And that's essentially why $H(C(A+B)) = H(C(A\cup B))$.

subdivided simplex

The precise proof is of course a bit more tedious, but that's essentially what's going on in Hatcher's Proposition 2.21.

Now we have a homomorphism $C(A) \oplus C(B) \to C(A + B)$ given by $(c_1, c_2) \mapsto c_1 + c_2$. This map is onto by definition of $C(A+B)$.

What is its kernel?

You should convince yourself that the kernel consists precisely of the elements of the form $(c,-c)$ with $c \in C(A \cap B)$. In other words, we have a short exact sequence of chain complexes:

$$0 \;\xrightarrow{\phantom{blablabla}} \; C(A \cap B) \; \xrightarrow[\phantom{blablabla}]{c \mapsto (c,-c)}\; C(A) \oplus C(B) \; \xrightarrow[\phantom{blablabla}]{(c_1,c_2) \mapsto c_1 + c_2}\; C(A+B)\; \xrightarrow[\phantom{blablabla}]{}\; 0$$

But this short exact sequence exhibits $C(A + B)$ as (isomorphic to) the quotient $\left(C(A) \oplus C(B)\right)\; / \; C(A\cap B)$. The Mayer-Vietoris sequence is then the long exact sequence in homology associated to this short exact sequence of complexes.

So the "difference" between $C(A + B)$ and $C(A) \oplus C(B)$ is precisely measured by $C(A \cap B)$: Chains lying in $A \cap B$ can belong to either $C(A)$ or $C(B)$. So no, $C(A+B)$ and $C(A) \oplus C(B)$ not the same sets.

A simple situation that shows intuitively what is going on here is: Let $X = \mathbb{R}^3$ and consider the subspaces $U = \langle e_1,e_2 \rangle$ and $V= \langle e_2, e_3 \rangle$ each spanned by two out of three vectors of the standard basis. Then $X = U + V$ while the (exterior) direct sum $U \oplus V$ has four dimensions. The superfluous dimension comes from the fact that $U \cap V = \langle e_2 \rangle$ is non-zero and we have a similar short exact sequence $0 \to U \cap V \to U \oplus V \to U + V \to 0$ as before.

share|improve this answer
    
Actually, I do draw pictures. I've been following your advice on more than one occasion... and I appreciate your help! I think I should chuck Hatcher, every now and again there is a passage in it that is utterly unhelpful : ( –  Matt N. Aug 9 '11 at 14:51
    
I wish I could give you more up votes for all this effort you put into writing this answer. –  Matt N. Aug 9 '11 at 14:54
    
@Matt: Very good, then. Look, you've been struggling with Hatcher for a long time now. Even if everybody likes that book, it need not be suitable for you. Maybe you should try some other book that suits your needs and inclinations a bit better? Maybe Massey, Munkres or Bredon? –  t.b. Aug 9 '11 at 14:54
2  
@Matt: Thanks, but it wasn't that bad. Drawing such pictures takes two or three minutes with GeoGebra. –  t.b. Aug 9 '11 at 14:56
    
GeoGebra looks really good, I had never heard of it before. I do all my drawing on paper, how old fashioned : D –  Matt N. Aug 10 '11 at 6:10
show 5 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.