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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.

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n-jective? :) ... – Blue Sep 3 '10 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. – Gunnar Þór Magnússon Nov 24 '10 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 – Willie Wong Nov 25 '10 at 0:09
What's wrong with non-injective? – Dan Christensen Jun 8 at 14:01

3 Answers 3

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

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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.

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+1: I think injective and surjective already confuse matters! (I always forget which is which :)) – Aryabhata Sep 3 '10 at 16:55
'sur' means on in French so surjective=onto is the way I remember it – WWright Sep 4 '10 at 16:38
An injective function injects one set into another, and a surjective function... doesn't. – Qiaochu Yuan Nov 24 '10 at 20:47

I will:

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


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:


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