Naturality/commutativity of Mayer-Vietoris giving wrong answer?

[nonequilibrium]

Hello,

I came across an argument for the fact that the degree of the map $R_n$ which reflects the n-sphere through a plane is -1. It goes as follows:

Describe $S^n$ as two disks whose overlap is $S^{n-1}$ (in such a way that $R_n$ restricted to this overlap is $R_{n-1}$)
Then due to naturality of the Mayer-Vietoris sequence, the following commutes:

$\begin{array}{ccc}<br /> H_n(S^n) &\to^\cong &H_{n-1}(S^{n-1}) \\<br /> \downarrow R & & \downarrow R \\<br /> H_n(S^n) &\to^\cong &H_{n-1}(S^{n-1})<br /> \end{array}$

Hence deg(R) is independent of the dimension n (and then we calculate deg(R) = -1 for $S^1$).

However: what if instead of the reflection $R$ we had used the inversion $\pi$? Every piece of the argument goes through (what would change?) but the conclusion would be wrong! (The degree of reflection depends on n, i.e. $\deg \pi = (-1)^{n+1}$.)

What is going wrong? Thanks! (and merry christmas)

 

[jgens]

The argument gives the correct value for the degree of a reflection, and further, the degree of the antipodal map is indeed (-1)ⁿ⁺¹. This can easily be seen by noting that the antipodal map on the n-sphere is a composition of n+1 reflections, and therefore, has degree (-1)ⁿ⁺¹. I have not checked why the argument above does not go through for the antipodal map, but I imagine it has something to do with a failure of commutativity. There are some conditions that need to hold on your map f:Sⁿ→Sⁿ in order to get commutativity of the relevant diagram.

 

[nonequilibrium]

I have not checked why the argument above does not go through for the antipodal map, but I imagine it has something to do with a failure of commutativity. There are some conditions that need to hold on your map f:Sn→Sn in order to get commutativity of the relevant diagram.

That is why I made this thread though: I can't find the condition that is not satisfied.

 

[jgens]

No guarantees on correctness here! Let X denote the n-sphere and consider the CW-triad (X,A,B) where A = {(x₁,...,x_n+1) | x_n+1 ≥ 0} and B = {(x₁,...,x_n+1) | x_n+1 ≤ 0} are n-cells whose common boundary is the (n-1)-sphere. Let f:X→X denote the reflection f(x₁,...,x_n+1) = (x₁,...,-x_n+1) and let g:X→X denote the antipodal map. It then follows that f(A) = g(A) = B and f(B) = g(B) = A. The idea is to compute the degrees of f and g using the Mayer-Vietoris sequence associated to the triad (X,A,B). Towards this end note there are two possible boundary maps ±∂_n(X)→H_n-1(A∩B) arising from the two natural short exact sequences 0→C(A∩B)→C(A)⊕C(B)→C(X)→0. The pushforwards f_#:C(X)→C(X) and g_#:C(X)→C(X) on chain complexes take the first of these short-exact sequences to the other. As a result we get the following two commutative diagrams:
$$\require{AMScd}<br /> \begin{CD}<br /> H_n(X) @>∂>> H_{n-1}(A \cap B)\\<br /> @Vf_*VV @V\mathrm{id}_*VV\\<br /> H_n(X) @>-∂>> H_{n-1}(A \cap B)<br /> \end{CD}<br /> \hspace{20mm}<br /> \begin{CD}<br /> H_n(X) @>∂>> H_{n-1}(A \cap B)\\<br /> @Vg_*VV @Vg_*VV\\<br /> H_n(X) @>-∂>> H_{n-1}(A \cap B)<br /> \end{CD}$$
From here the degrees are easily determined. This argument is a little different than the one outlined in your OP. To be completely honest I am not entirely sure how that calculation goes through. But this new computation works formally and gives the correct result for both the reflection and antipodal maps.

 

[nonequilibrium]

That seems to work :)

Your explanation also works for the one in the OP: there the plane of reflection is just chosen different (say $x_1 \to - x_1$) which has two consequences:

(1) f(A) = A and f(B) = B,
(2) the reflection on $S^{n+1}$ restricts to the identity on the reflection on $S^n$. (Instead of the identity as in your choice.)

This gives the commutative diagram

$$\require{AMScd}<br /> \begin{CD}<br /> H_n(X) @>∂>> H_{n-1}(A \cap B)\\<br /> @Vf_*VV @V f_*VV\\<br /> H_n(X) @>∂>> H_{n-1}(A \cap B)<br /> \end{CD}$$

which is the one stated in the OP. If we look at inversion instead of reflection, again f(A) = B and f(B) = A such that on the bottom $\partial \to - \partial$. This gives all the right results :)

Thanks! (Note your solution is nicer than the one in the OP since there induction was used and the degree for reflection on the 1-sphere had to be calculated explicitly as the base case. But in that case the subtlety of commutativity could be swept under the rug, whereas in your case we are forced to think about the minus signs (which is interesting))