What is the domain of #(f@g)(x)#?

1 Answer
Jul 24, 2015

If #g:A->B# and #f:B->C#, then the domain of #f@g# is

#bar(g)^(-1)@bar(f)^(-1)(C)#

using the notation described below...

Explanation:

If #g# is a function that maps some elements of a set #A# to elements of a set #B#, then the domain of #g# is the subset of #A# for which #g(a)# is defined.

More formally:

#g sube A xx B :#

#AA a in A AA b_1, b_2 in B#

#((a, b_1) in g ^^ (a, b_2) in g) => b_1 = b_2#

Use the notation #2^A# to represent the set of subsets of #A# and #2^B# the set of subsets of #B#.

Then we can define the pre-image function:

#bar(g)^(-1): 2^B -> 2^A# by #bar(g)^(-1)(B_1) = {a in A : g(a) in B_1}#

Then the domain of #g# is simply #bar(g)^(-1)(B)#

If #f# is a function that maps some elements of set #B# to elements of a set #C#, then:

#bar(f)^(-1): 2^C -> 2^B# is defined by #bar(f)^(-1)(C_1) = {b in B : f(b) in C_1}#

Using this notation, the domain of #f@g# is simply

#bar(g)^(-1)(bar(f)^(-1)(C)) = (bar(g)^(-1)@bar(f)^(-1))(C)#