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I just want to make sure this is all correct.

So my definition of a function $f:A\to B$ being a surjection is:

For all $b \in B$, there exists an $a \in A$ such that $f(a) = b$.

Now the question:

Let $f: \mathbb N \to \mathbb Z \\ n \mapsto -n$.

Then $f$ is not surjective.

Proof:

Take an arbitrary $b \in \mathbb Z$. Let $a = -b \geq 0$.

We see that $f(a) = b$ but not every $a$ can be $-b$, because if we have an positive $b$ then $a$ cant be negative. (Since $a$ is in the natural numbers).

However: What is the function was $f:\mathbb Z \to \mathbb Z$. This would be a surjection, right?

flawr
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k9b
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2 Answers2

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For your second question, yes, it is a surjection.

But since you want to show that your first $f$ isn't a surjection, it's better to exhibit a specific $b \in \mathbb{Z}$ which is not in the range of $f$. For example, show that there is no $a \in \mathbb{N}$ such that $f(a) = 1$.

I don't think what you've written is very clear. When you say "Let $a = -b$," what you really need to point out is that $a = -b$ is the only $a$ for which you could possibly have $f(a) = b$. (I know this is obvious, but it's the main logical step in your proof.) Then show that in some cases $-b$ doesn't belong to $\mathbb{N}$.

user191371
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  • Very unclear indeed. But this is the way it always is when we’re just beginning to understand. One of the big challenges in mathematics is to improve our understanding so that we can explain with clarity. – Lubin Nov 10 '14 at 19:09
  • Thanks for the clarifications and I will try to be more clear when proving – k9b Nov 10 '14 at 19:52
  • @k9b I didn't mean that as a criticism of your effort. Your clarity will improve with experience in reading and writing proofs. – user191371 Nov 11 '14 at 08:25
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Yes it is, since $f(x) = -x$ is bijective on $\mathbb Z$.

flawr
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