# A $1-1$ function is called injective. What is an $n-1$ function called?

A $1-1$ function is called injective. What is an $n-1$ function called ?

I'm thinking about homomorphisms. So perhaps homojective ?

Onto is surjective. $1-1$ and onto is bijective.

What about n-1 and onto ? Projective ? Polyjective ?

I think $n-m$ and onto should be hyperjective as in hypergroups.

• n-jective? :) ...
– Blue
Sep 3, 2010 at 7:17
• These aren't called anything in particular, but I've seen algebraic geometers denote that a function is $n$ to $1$ by writing $n:1$ above an arrow between sets. You have to keep in mind that these functions usually arise either as finite covering maps or polynomials (or both at the same time), so it's usually clear when one actually has to use that a given function has that property. Nov 24, 2010 at 21:59
• A slightly related concept from complex function theory and the theory of automorphic functions is valence. A holomorphic function on an open domain of the complex plane is said to be univalent if it is 1-to-1, and $p$-valent if it is at most $p$-to-1. See also eom.springer.de/m/m065560.htm Nov 25, 2010 at 0:09
• What's wrong with non-injective? Jun 8, 2015 at 14:01

n-1 + onto is sometimes called n-fold cover (by analogy).

IMHO, an n to 1 function should be called "an n to 1 function." Simple, decriptive, to the point. Adding a new term in this case just muddies the waters.

• +1: I think injective and surjective already confuse matters! (I always forget which is which :)) Sep 3, 2010 at 16:55
• 'sur' means on in French so surjective=onto is the way I remember it Sep 4, 2010 at 16:38
• An injective function injects one set into another, and a surjective function... doesn't. Nov 24, 2010 at 20:47

I will:

• suggest some terminology for three related concepts, and
• suggest that $n$-to-$1$ functions probably aren't very interesting.

Terminology.

Let $f : X \rightarrow Y$ denote a function. Recall that $f$ is called a bijection iff for all $y \in Y$, the set $f^{-1}(y)$ has precisely $1$ element. So define that $f$ is a $k$-bijection iff for all $y \in Y$, the set $f^{-1}(y)$ has precisely $k$ elements.

We have:

The composite of a $j$-bijection and a $k$-bijection is a $(j \times k)$-bijection.

There is also a sensible notion of $k$-injection. Recall that $f$ is called an injection iff for all $y \in Y$, the set $f^{-1}(y)$ has at most $1$ element. So define that $f$ is a $k$-injection iff for all $y \in Y$, the set $f^{-1}(y)$ has at most $k$ elements.

We have:

The composite of a $j$-injection and a $k$-injection is a $(j \times k)$-injection.

There is also a sensible notion of $k$-subjection, obtained by replacing "at most" with "at least."

We have:

The composite of a $j$-surjection and a $k$-surjection is a $(j \times k)$-surjection.

A criticism.

I wouldn't advise thinking too hard about "$k$ to $1$ functions." There's a couple of reasons for this:

1. Their definition is kind of arbitrary: we require that $f^{-1}(y)$ has either $k$ elements, or $0$ elements. Um, what?

2. We can't say much about their composites: Perhaps these types of $$n-1$$ functions are called "$$n-to-1$$" functions. There is also a "$$Division$$ $$Rule$$" which states that if $$f$$ is a $$n-1$$ function, then $$|X| =n|Y|$$, in set theory and counting principles.