I have a question here :

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Now the way I have been taught to prove surjectivity is as follows : Let $y$ = $h(x)$ Hence $y=x+5$ $\Rightarrow$ $y-5=x$ $\Rightarrow$ $h(x) = h(y-5) = y-5+5 = y$

and hence I can say that the function is a surjection.

Now. I know that since the co-domain has to be in $\Bbb N$ we have a case of $x+5=1$ and for this to work $x$ has to be $-6$, and the domain must also be in $\Bbb N$ but $-6\notin\Bbb N$.

So how does this work? How do I prove surjectivity? How did It workout the first time even though a single example disproved the entire thing?



1 Answer 1


A function $f: A \to B$ is surjective if and only if there exists at least one $a \in A$ corresponding to all $b \in B$, that is $f(a) = b$. Generally, we prove (or disprove) surjectivity by using this definition.

Now, in your case, notice that we use this definition in order to show that the function is not surjective. We have $h: \mathbb{N} \to \mathbb{N}$ and since there doesn't exist some $n \in \mathbb{N}$ corresponded to $0 \in \mathbb{N}$, in other words, there doesn't exist $n \in \mathbb{N}$ such that $f(n) = 0$, so we can conclude that the function is not surjective.

  • $\begingroup$ 0 $\notin\Bbb N$ though? $\endgroup$
    – K.M.
    Jan 9, 2018 at 18:54
  • $\begingroup$ I'm with you on the fact that it can be proven to be surjective by finding a coounter example, however my question is why does the first method I used prove it to BE surjective? $\endgroup$
    – K.M.
    Jan 9, 2018 at 18:56
  • $\begingroup$ Codomain and range are two different things. Here, codomain is $\mathbb{N}$ and includes $0$ while the range is $\{5,6,7,...\} = \mathbb{N} - \{0,1,2,3,4\}$. In surjections, range must be equal to codomain. $\endgroup$
    – ArsenBerk
    Jan 9, 2018 at 18:57
  • $\begingroup$ Can you explain the first method you used more? Because i don't understand why $h(x) = h(y-5) = y-5+5 = y$ implies $h$ is a surjection. This equality just gives you $h(x) = y$ which has nothing to do with surjectivity. Or it can imply $h(y-5) = y$, which is still not a surjection because $y$ is the range of $h$, which is $\{5,6,7,...\}$. $\endgroup$
    – ArsenBerk
    Jan 9, 2018 at 19:03
  • 1
    $\begingroup$ Yes, it is kinda vicious cycle. Definitions are always good to use :) $\endgroup$
    – ArsenBerk
    Jan 9, 2018 at 19:20

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