Does there exist an injective function that is not surjective? Could I have an example, please?

  • $\begingroup$ What is the definition of surjective according to you? $\endgroup$ – imranfat Oct 30 '13 at 20:01
  • $\begingroup$ @imranfat The function $\operatorname{f} : U \to V$ is surjective if for each $v \in V$, there exists a $u\in U$ for which $\operatorname{f}(u)=v$. This is a standard definition. $\endgroup$ – Fly by Night Oct 30 '13 at 20:08
  • $\begingroup$ In future, you should give us more background on what you know and what you have thought about / tried before just asking for an answer. $\endgroup$ – Ahaan S. Rungta Nov 1 '13 at 16:14
  • $\begingroup$ See also: math.stackexchange.com/questions/991894/… $\endgroup$ – Martin Sleziak Oct 26 '14 at 16:18

There are many examples. It just all depends on how your define the range and domain.

For example $\operatorname{f} : \mathbb{R} \to \mathbb{R}$ given by $\operatorname{f}(x)=x^3$ is both injective and surjective. But then I can change the image and say that $\operatorname{f} : \mathbb{R} \to \mathbb{C}$ is given by $\operatorname{f}(x) = x^3$. Now it is still injective but fails to be surjective.

Similarily, the function $\operatorname{g} : \mathbb{R} \to \mathbb{R}$ given by $\operatorname{g}(x)=x^2$ is neither surjective nor injective. But if I change the range and domain to $\operatorname{g}: \mathbb{R}^+ \to \mathbb{R}^+$ then it is both injective and surjective.

Edit: As requested by the OP:

An example of an injective function $\mathbb{R}\to\mathbb{R}$ that is not surjective is $\operatorname{h}(x)=\operatorname{e}^x$. This "hits" all of the positive reals, but misses zero and all of the negative reals. But the key point is the the definitions of injective and surjective depend almost completely on the choice of range and domain.

  • $\begingroup$ Thank you for example $\operatorname{f} : \mathbb{R} \to \mathbb{C}$. And an example of injective function $\operatorname{f} : \mathbb{R} \to \mathbb{R}$ that is not surjective? $\endgroup$ – Mark Nov 1 '13 at 16:00
  • $\begingroup$ @Mark See above. $\endgroup$ – Fly by Night Nov 1 '13 at 16:13
  • $\begingroup$ Take any bijective function $f:A \to B$ and then make $B$ "bigger". $\endgroup$ – steven gregory May 16 at 1:28

How about $f(x)=e^x.$ Your job is to figure out the domain and range. You may want to use the fact that strictly monotone functions are injective.

  • $\begingroup$ Surjective means that every "B" has at least one matching "A" So B is range and A is domain. Why isn't the e-power function surjective then? $\endgroup$ – imranfat Oct 30 '13 at 20:04
  • $\begingroup$ @imranfat It depends completely on the range and domain. Without those, the words "surjective" and "injective" have no meaning. $\endgroup$ – Fly by Night Oct 30 '13 at 20:05
  • $\begingroup$ I guess that makes sense. I like the one-to-one idea much more. We use it with inverses and transcendental functions in Calc. I learned about terms like surjective, injective and bijective so long ago, it seems like these terms aren't so popular anymore $\endgroup$ – imranfat Oct 30 '13 at 22:53
  • 1
    $\begingroup$ Yes, surjective is kind of weird like that. But in questions that come up, usually there are two spaces we start with then we want to see if a function from one to the other is surjective, and it may not be easy. Every function is surjective onto its image but this does not help with many problems. $\endgroup$ – Maxim G. Oct 31 '13 at 0:54

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