Domain and codomain for y =[tex]\sqrt{1-x^2}[/tex]

[foxtrot_echo_]

Consider the mapping f: X$$\rightarrow$$Y where f(x)=y=$$\sqrt{1-x^2}$$

consider the co-domain Y , we can define the mapping over [-1,1] $$\rightarrow$$ $$\mathbb R$$ , ( in this case the mapping won't be onto)
and in case we define the mapping over [-1,1] $$\rightarrow$$ [0,1] (in this case mapping is onto)

(is my understanding till this point right?)

and my question is
does it make any sense to say that the domain X is the set $$\mathbb R$$ or can the mapping only be defined such that X is the set [-1,1] (or its subsets) ?
(note I am not considering case of complex numbers)

and another minor quibble :- when we plot y = $$\sqrt{1-x^2}$$ , why do we show the domain X over entire real line shouldn't we only show the line segment from [-1,1] (if it doesn't make any sense to say X is entire $$\mathbb R$$) ?

 

[foxtrot_echo_]

basically , my confusion is this :-
in functions which are not 'onto' , we are not pairing all the elements of the co-domain with elements of domain.

but does the opposite make sense , i.e can we speak of a domain where some elements are not paired with elements of co-domain?

Sorry if question is very basic,please bear with me over this.