# Confusion between function and multivalued function.

"What is a function?" can be answered as "Single-valued relations are called functions". But how can "What are the multi-valued function?" be answered?

Will someone clarify my doubt why multi-valued functions are not violating the classical definition of function?

EDIT

This is what Wikipedia says on multivalued functions:

In mathematics, a multivalued function from a domain X to a codomain Y is a heterogeneous relation. However, in some contexts such as the complex plane (X = Y = ℂ), authors prefer to mimic function theory as they extend concepts of the ordinary (single-valued) functions.

In this context, an ordinary function is often called a single-valued function to avoid confusion.

The term multivalued function originated in complex analysis, from analytic continuation. It often occurs that one knows the value of a complex analytic function $f(z)$ in some neighbourhood of a point $z=a$. This is the case for functions defined by the implicit function theorem or by a Taylor series around $z=a$. In such a situation, one may extend the domain of the single-valued function $f(z)$ along curves in the complex plane starting at $a$. In doing so, one finds that the value of the extended function at a point $z=b$ depends on the chosen curve from $a$ to $b$; since none of the new values is more natural than the others, all of them are incorporated into a multivalued function. For example, let $f(z)=\sqrt {z}$, be the usual square root function on positive real numbers. One may extend its domain to a neighbourhood of $z=1$ in the complex plane, and then further along curves starting at $z=1$, so that the values along a given curve vary continuously from $\sqrt {1}=1$. Extending to negative real numbers, one gets two opposite values of the square root such as $\sqrt {-1}=\pm i$, depending on whether the domain has been extended through the upper or the lower half of the complex plane. This phenomenon is very frequent, occurring for $n$th roots, logarithms and inverse trigonometric functions.

To define a single-valued function from a complex multivalued function, one may distinguish one of the multiple values as the principal value, producing a single-valued function on the whole plane which is discontinuous along certain boundary curves. Alternatively, dealing with the multivalued function allows having something that is everywhere continuous, at the cost of possible value changes when one follows a closed path (monodromy). These problems are resolved in the theory of Riemann surfaces: to consider a multivalued function ${\displaystyle f(z)}$ as an ordinary function without discarding any values, one multiplies the domain into a many-layered covering space, a manifold which is the Riemann surface associated to $f(z)$.

• They are violating the definition of function. – Lord Shark the Unknown Aug 10 '18 at 18:07