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

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The mapping f: X → Y defined by f(x) = y = √(1-x²) has a domain of X that can only be defined as [-1, 1] or its subsets, as values outside this range do not yield real outputs. When the mapping is defined over [-1, 1] to [0, 1], it is onto, meaning every element in the codomain is paired with an element in the domain. Plotting y = √(1-x²) over the entire real line is misleading since the function is only valid within the restricted domain of [-1, 1]. It is acceptable to discuss a domain where not all elements are paired with elements of the codomain, as the domain can include values that do not produce valid outputs. Understanding these mappings clarifies the relationship between domain and codomain in mathematical functions.
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Consider the mapping f: X\rightarrowY 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) ?
 
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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.
 
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