# HW3 Problem 1(b)

In the problem we are to show that $(2-\zeta )^{-1}[x,y] = number\; of\;chains \;between \;x \;and\; y$. Is the ‘2’ here the constant function that sends everything to 2 or what is it? Thanks.

1. Yes, in this class, “What is 2?” is a great question.

It is $2 \cdot \mathbf{1}$ where 2 is the number 2 in the field, and $\mathbf{1}$ is the multiplicative identity of the incidence algebra, which sends the intervals $[x,x]$ to 1 and all other intervals to 0.

2. sammonkie

Silly thought:

Let

$f=(2-\zeta )$ and $k=\# \;chains \;between \;x \;and\; y$. So isn’t what we are to prove equivalent to
$1_A[x,y] =f\;f^{-1}[x,y] = k(f\;1_A)[x,x]$?

I know something is wrong here but What is going wrong? Thanks.

3. sammonkie

Never mind… that is a very silly mistake I made indeed… -_- (kids,don’t try to do it at home! =D)