On 10/06/2013 04:42 PM, Peter Percival wrote: > Arturo Magidin wrote: > >> >> Technically, two functions are equal if and only if they have the >> same domain, and they take the same value at each point on the >> domain. > > I wonder why you don't require the codomains to be the same as well. > Two functions might be equal according to that definition but one could > be invertible and the other not. >
In the way I think of a function 'f', no co-domain attaches to 'f' in a natural way. For example, for the function f with domain R (= the set of real numbers) and such that, for any x in R,
f(x) = sin(x),
the domain of f is R, but I would write: "f with co-domain R", "f with co-domain [-1, 1]", or by abuse of language, "f(x) with co-domain R", "f(x) with co-domain [-1,1]".
Nevertheless, the default, primary, definition of "function" at PlanetMath provides for a co-domain being attached to a function. They say that "in set theory", a function is a special type of relation, so along the line of thought in Arturo Magidin's way of thinking or my way of thinking.
The "no co-domain included" definition is convenient, because for example we can say
f: C -> C is the polynomial function defined by f(z):= z^137 + 202 z^23 - 45z^4 + 32 z^3 - 1001 z^2 + 7z , all this without bothering to specify a co-domain.
And in any case, if we say "f maps C onto C", it's equivalent to saying f(C) = C.