Codomains of composite functions

[airborne18]

Homework Statement

Hopefully simple. Do composite functions have to have the same Codomain? What if they do not, does the smaller Codomain get canceled out? f(x) : R ->R g(x) :Z->Z f(x) g(x) : R->R Is this correct? Or do I need to hit the books a bit more?

Homework Equations

The Attempt at a Solution

 

[Office_Shredder]

The codomain of f(x) and g(x) are basically unrelated. If you want to define f(g(x)), the key aspect is that the codomain of g(x) and the domain of f(x) line up. If not, there will be values of x for which g(x) can't be plugged into f(x) and you fail to have a function. In such a case you have to restrict the domain of g(x) so that the codomain of g(x) is a subset of the domain of f(x)

Also, assuming that f(x) g(x) in your post means f(g(x)) (so we have function composition), then the domain has to be only the integers: you can't plug an arbitrary real number into g(x). Actually this is true even if you meant multiply the two functions

 

[airborne18]

The codomain of f(x) and g(x) are basically unrelated. If you want to define f(g(x)), the key aspect is that the codomain of g(x) and the domain of f(x) line up. If not, there will be values of x for which g(x) can't be plugged into f(x) and you fail to have a function. In such a case you have to restrict the domain of g(x) so that the codomain of g(x) is a subset of the domain of f(x)

Also, assuming that f(x) g(x) in your post means f(g(x)) (so we have function composition), then the domain has to be only the integers: you can't plug an arbitrary real number into g(x). Actually this is true even if you meant multiply the two functions

Thanks. My assumption was that if there were values that were not valid for g(x), the function would simply be undefined for those values in R that are inconsistent with Z. That is what is screwing me up with this. My problem is that I always put Codomain = Range, and that has been a flaw in my thinking as well.