# Surjectivity and multivalued functions

I have a question with regards to surjectivity and multivalued functions. If I have a domain with one element and a mapping to the codomain, in which the codomain consists of three elements and that element in the domain is mapped to all the three distinct elements in the codomain, then is the mapping surjective? Because for all elements in the codomain there is one in the domain which maps to it. But I haven't seen that as an example of surjectivity, may someone explain to me why? or is it because I just haven't found an example yet?

## 1 Answer

It's not a function that you defined but a relation which is total on the codomain.

A relation (or multivalued partial function, if you prefer) $$A\to B$$ is a function, by definition, iff it is single-valued and total on the domain, meaning that every $$a\in A$$ is in relation with a unique $$b\in B$$.

The term 'surjective' is usually applied only to functions.

• Oh I see, so a mapping F: A-->B is a function if it maps every element of A to a unique element in the set B. Just to clarify, so the relation I was discussing is in a sense "surjective' since for all y in the codomain there is a unique element in the domain such that f(a)=b, but it is not a function since it doesn't follow the definition that it maps every element in A to a unique element in B. Am I correct? – user612135 Nov 23 '18 at 1:59
• Yes, almost completely right, but surjectivity doesn't imply uniqueness of preimage. Also, the terms 'mapping', 'map' are all used as synonyms for 'function'. Simply use 'relation' (or add adjective 'multivalued') instead, for clarity. – Berci Nov 23 '18 at 8:49
• I see, that was of great help thanks! – user612135 Nov 23 '18 at 22:15