A function is a relation for elements in a to elements in b such that for every element in a there is a unique element in b. This is vacuously true of the empty relation when a is empty.

I didnt discuss a function, I discussed a mapping. They are different constructs. Imagine the question like this: you have two groups of people, students and teachers. How many possible ways are there to assign students to teachers? To put it another way, how many arrows would it take pointing from the student to the teacher to illustrate every possible assignment? Now, if there are no students and no teachers, you have no arrows, not one.

So mapping is exactly a function, unless domain and codomain are empty, in which case it doesn't exist?

Empty set is as good and useful as nonempty ones, and you seen to have no problems with using them. Why do you find it hard to accept an empty function f: E -> X, where E is empty set, that doesn't associate any element of E with any element of X?

I am a mathematician I don't need examples. What is your formal definition of a mapping.

In mathematic it is normally used as a synonym for a morphism in a given category and in the category of sets this would be a function. So in our context a mapping is a function.

You're being obtuse. If you read what he was saying, it would be obvious that he didn't mean a formalized "mapping" and meant a binary relation. Instead of using this as an opportunity to evangelize maths and teach, you criticized him with a pretentious self-aggrandizing glory that Mathematicians already have such a bad reputation for.

> I am a mathematician I don't need examples.

Falser words were never spoken.

> I am a mathematician I don't need examples.

I gave the example not to teach you, but so that we could get away from terminology and get to meaning. Look at the question I posed: "how many arrows would it take pointing from the student to the teacher to illustrate every possible assignment?" So, if you have two students Alice and Eve, and two teachers Bob and Carl, there are four arrows: Alice->Bob, Alice->Carl, Eve->Bob, and Eve->Carl. How many arrows are there if there are no teachers and no students?

The number of arrows is |students x teachers|, every student paired with every teacher. From this you get a^0 = 0, but it also follows that a^b = b^a. This defines multiplication, not exponentiation. The problem is that you cannot illustrate the assigment where Eve is not paired with Bob using an arrow, but only by not using a one.