I don't fully understand surjective proofs.

Here is a baby example I worked on.

PROPOSITION: Let f $\mathbf{R} \rightarrow \mathbf{R}$ be the function defined as $f(n)= 2n+3$.

Proof f is surjective: For $f$ to be surjective our range, must be in all of $\mathbf{R}$. Let's assume y$ \in \mathbf{R}$, then let's take the inverse of your function to attain $n=(y-3)/2$. We can see $y, n \in \mathbf{R}$, but more importantly we can see $n$ in the domain. Note if we plug this back into our function we would attain $$ 2(y-3)/2+3$$ $$ = (y-3)+3$$ $$ = y$$ This shows that $f$ is surjective.

I think my assumption is off. I know I have to prove directly. I believe that means we can attain a y for from a given n. Therefore we must assume y in $\mathbf{R}$ then.... I am a little lost. My guess is that we would use $f(n)=$ some way to find $f(n)=\mathbf{R}=y$, but there is something I missing.

Thank you for your response.

  • $\begingroup$ If $f(n)=2n+3$ then $f(\mbox{input})=f((y-3)/2)=y$. But you have to get the "input" in some way. For example, as you have done. $\endgroup$
    – mfl
    May 19, 2021 at 19:22
  • $\begingroup$ You take any $y$ in the codomain and try to find an $n$ in the domain that maps to $y$; for this you have to find out $n$ in terms of $y$ by "undoing the equation" $y=f(n)$ which expresses $y$ as a function of $n$, and verify that $f(n(y))$ is indeed $y$. $\endgroup$ May 19, 2021 at 19:25
  • $\begingroup$ Your proof is perfectly fine. When you want to prove a function $f:A\rightarrow B$ is surjective (or onto), you need to prove that for every $y\in B$ there is at least one $x\in A$ such that $f(x)=y$, so you start with a generic $y$ and construct (like you did) at least one $x\in A$ (there might be more than one such $x$) that is mapped to $y$. Notice that the range $B$ is important, if you consider a larger $B'\supseteq B$ as range, the function $f$ may not be surjective anymore $\endgroup$
    – Alessandro
    May 19, 2021 at 19:26

2 Answers 2


Your assumption is not that off. In order to prove that a function is surjective you can either prove this directly. Or you can find the inverse function proving that the function is bijective(just like you have done), thus proving the function is surjective.

Proving $\displaystyle f(x) =2+3$ is surjective directly:

Let $\displaystyle x\in \mathbb{R}$ We need to prove that there exists an $\displaystyle x$ S.T- $\displaystyle f( x) =y$

We choose $\displaystyle x=\frac{1}{2} y-\frac{3}{2}$

Which follows-

$\displaystyle \ \ \ \ \ f( x) =2x+3\rightarrow \ \ f\left(\frac{1}{2} y-\frac{3}{2}\right) =2\cdotp \left(\frac{1}{2} y-\frac{3}{2}\right) +3=y$


The reasonning you used here is known as "Analysis-Synthesis" , which means that, for example, if you want to prove that a cetain proposition $ P $ is true, you do it in two steps :


You begin by assuming that $ P $ is true and you find a necessary condition $ Q $, then

  1. Synthesis

You check that this necessary condition $ Q $ , is sufficient.

In your case : Given $ y\in \Bbb R$, you want to prove that

$$P : (\exists x\in \Bbb R)\;:\; f(x)=y$$

You got the necessary condition $$Q : x=\frac{y-3}{2}$$

  • $\begingroup$ Isn't it what Esty.R has just proved? have a look... $\endgroup$
    – Chopin
    May 19, 2021 at 19:49
  • $\begingroup$ @aasc232 He didn't specify the kind of reasonning used as asked by the OP. $\endgroup$ May 19, 2021 at 19:53

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