Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'd would like to know that, because I don't want to prove a function is onto if I don't have to. If the answer is no, is there any particular case where it is true?

share|improve this question
    
Any function is onto its definition domain's image (codomain), so any 1-1 function is bijective with its codomain. –  DonAntonio Oct 28 '12 at 19:39
2  
@DonAntonio In my experience, "codomain" is used to denote the set into which the function maps, while "range" is used to denote the image of the function. –  Austin Mohr Oct 28 '12 at 19:42
    
@AustinMohr, you may well be right. I really am not sure of this term. –  DonAntonio Oct 28 '12 at 19:47

4 Answers 4

up vote 12 down vote accepted

The function $f:\Bbb N\to\Bbb N:n\mapsto 2n$ is one-to-one but not onto, and there are many other easy examples. An even simpler one is $g:\Bbb N\to\Bbb N:n\mapsto n+1$.

About the only simple situation in which a one-to-one function $f:A\to B$ is necessarily onto is when $A$ and $B$ are finite sets of the same cardinality.

share|improve this answer
    
(+1) for giving a nice sufficient condition. –  Austin Mohr Oct 28 '12 at 19:43

The answer is no in general, although it does depend on the specific situation. The exponential function is one-to-one but it is not onto if we consider the co-domain to be $\mathbb{R}$. It is onto if we further restrict the co-domain to $\mathbb{R}^+$.

One prominent case in which one-to-one implies onto (and vice versa) is for linear mappings between finite dimensional vector spaces. If we have a linear mapping between spaces of equal dimension, then the mapping is one-to-one if and only if it is onto.

share|improve this answer
    
You cannot conclude that an injective linear map between two vector spaces of the same dimension is necessarily surjective without assuming your spaces are finite dimensional. Consider the map $(x_1,x_2,\dotsc) \mapsto (0,x_1,x_2,\dotsc)$ on some sequence space. –  kahen Oct 29 '12 at 3:24
    
@kahen Yes you are right. I've added the restriction. –  EuYu Oct 29 '12 at 3:33

The function $f: \{0\} \rightarrow \mathbb{R}$ defined by $f(x) = x$ is one-to-one but not onto. (Don't be afraid to be creative with your domains and codomains!)

For a less "gimmicky" example, $f: \mathbb{R}^+ \rightarrow \mathbb{R}$ defined by $f(x) = x^2$ is one-to-one but not onto (the square of a real number is never negative).

share|improve this answer
    
Your second example does not have the indicated domain (or the indicated codomain). –  Andres Caicedo Oct 28 '12 at 21:50
    
@AndresCaicedo Thank you for noticing my oversight. The example I have now is what I originally intended. –  Austin Mohr Oct 28 '12 at 22:35

A function is one to one if it has an inverse function (the horizontal line check), that is for all values $x\neq x'$, then $f(x)\neq f(x')$ :).

share|improve this answer
2  
The statement is a bit problematic. Injective functions are not equivalent to invertible functions (although they admit an inverse when restricting the co-domain to the image). –  EuYu Oct 28 '12 at 22:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.