HW 3, Exercise 1, part b: Is (f+g)[x, y]=f[x,y]+g[x,y]?

I wanted to point this out and see if this is valid:

Since 2[x, y]=2*1[x, y]=0 and \zeta[x,y]=1, does that mean that 2-\zeta[x, y]=2[x,y]-\zeta[x,y]=0-1=-1?

More generally, is it true that (f+g)[x,y]=f[x,y]+g[x,y]?

I am inclined to think that this is the case, but maybe the way the multiplication is, i.e., ff'[x,y]=\sum f[x,z]f'[x,y], bars us from claiming so.

An example would be 2[x,y]=1[x,y]+1[x,y], where 1[x,y] is as is in the problem statement. This clearly works, but does it work in general for functions f, g?

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About El Niño

I'm a math geek. Math is fun!

4 comments

  1. sammonkie

    I think so — the functions are in Hom which is a vector space.

  2. sammonkie

    Yes — they are evaluation maps: +  =

  3. sammonkie

    Yes — they are evaluation maps: $latex \ + \ = \$

  4. Wonderful! Well, based on the way the multiplication works on A(P), the above result will very likely be of little consequence to the proof of the exercise, though it is still necessary.

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