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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?

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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
@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
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.

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(+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.

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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).

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Your second example does not have the indicated domain (or the indicated codomain). – Andrés E. 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

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