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.

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3 comments

  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)

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